Drorbn - User contributions [en]

10-327/Homework Assignment 6

2010-11-17T23:54:40Z

Ian: /* Due date */

{{10-327/Navigation}}
===Reading===
'''Read''' sections 37-38 in Munkres' textbook (Topology, 2nd edition). Remember that reading math isn't like reading a novel! If you read a novel and miss a few details most likely you'll still understand the novel. But if you miss a few details in a math text, often you'll miss everything that follows. So reading math takes reading and rereading and rerereading and a lot of thought about what you've read. Also, '''preread''' sections 30-33, just to get a feel for the future.

===Doing===
Solve and submit the following problems.

'''Problem 1.''' Problem 1 on page 235 of Munkres' book. (Though following the [[10-327/Errata_to_Munkres'_Book|Errata to Munkres' Book]], in part (c) replace <math>T_1</math> by <math>T_2</math>).

'''Problem 2.''' Show that <math>\{0,1\}^{\mathbb N}</math> is homeomorphic to the cantor set <math>C</math>.

'''Problem 3.''' Show that any function <math>f\colon{\mathbb N}\to I^A</math> from the integers into a "cube" <math>I^A=[0,1]^A</math> has a unique continuous extension to <math>\beta{\mathbb N}</math>.

'''Problem 4.''' Use the fact that there is a countable dense subset within <math>I^I</math> to show that the cardinality of <math>\beta{\mathbb N}</math> is greater than or equal to the cardinality of <math>I^I</math>.

'''Problem 5.''' Show that the cardinality of <math>\beta{\mathbb N}</math> is also less than or equal to the cardinality of <math>I^I</math>, and therefore it is equal to the cardinality of <math>I^I</math>.

'''Problem 6.''' Show that if <math>\mu\in\beta{\mathbb N}\backslash{\mathbb N}</math> and if <math>\mbox{Lim}_\mu</math> is the corresponding generalized limit, and if <math>b</math> is a bounded sequence and <math>f\colon{\mathbb R}\to{\mathbb R}</math> is a continuous function, then <math>\mbox{Lim}_\mu f(b_k) = f(\mbox{Lim}_\mu b_k)</math>.

'''Problem 7.''' Show that there is no super-limit function <math>\mbox{SuperLim}</math> defined on bounded sequences of reals with values in the reals which has the following 4 properties:
# <math>\mbox{SuperLim}(a_k)=\lim a_k</math>, if the sequence <math>a_k</math> is convergent.
# <math>\mbox{SuperLim}(a_k+b_k) = \mbox{SuperLim}(a_k) + \mbox{SuperLim}(b_k)</math>.
# <math>\mbox{SuperLim}(a_k\cdot b_k) = \mbox{SuperLim}(a_k) \cdot \mbox{SuperLim}(b_k)</math>.
# <math>\mbox{SuperLim}(a'_k) = \mbox{SuperLim}(a_k)</math>, where <math>a'</math> is <math>a</math> "shifted once": <math>a'_k=a_{k+1}</math>.

===Due date===
This assignment is due at the end of class on Thursday, November 18, 2010.

{{Template:10-327:Dror/Students Divider}}

Hi Dror, how do we prove some set has smaller cardinality to another set? I mean what do you mean by that?
Why in problem 5 it says that the cardinality of a set is less or equal to another set so that the cardinality are the same?
[[User:Xwbdsb|Xwbdsb]] 00:25, 13 November 2010 (EST)
*One way to do this is that if you have a map <math>f(x):A \to B</math> such that f(x) is 1-1, then <math>A</math> has cardinality less than or equal to <math>B</math>. I haven't worked on this question yet, so it may be a red herring, but in general I find the technique useful. [[User:Johnfleming|John]]

Also I am wondering what is the super-limit because it wasn't done in class and it is not covered in the book too??[[User:Xwbdsb|Xwbdsb]] 00:36, 13 November 2010 (EST)
I remember that when proving the homeomorphism in question 2 there is another intricate argument for continuity that is
not the standard definition in our topology class? Sorry Dror I am sort of lost of what to do for this assignment...[[User:Xwbdsb|Xwbdsb]] 00:36, 13 November 2010 (EST)

Also what is the generalized limit? I search for this idea in the book but I didn't find anything. I think in the book before we understand Stone-Cech compatification theorem we need to understand what a regular space is and also we need to understand one-point compactification? Isn't Stone-Cech compatification just a special way to compatify the some topological space so that the continuous function with uniquely be extended to the compatification? [[User:Xwbdsb|Xwbdsb]] 00:58, 13 November 2010 (EST)

* I actually genuinely enjoyed writing this assignment, thinking that I've asked a lovely collection of questions that will challenge you in the positive sense of the word - make you scratch your heads, and struggle a bit, and fight a bit to understand what is going on and eventually get some satisfaction and gain some appreciation of the intricacy and beauty of the subject matter as you understand and solve these problems. So please challenge yourself and think and struggle a bit more. [[User:Drorbn|Drorbn]] 10:32, 13 November 2010 (EST)

* I have a question concerning Problem 7. Perhaps I am misunderstanding it, but if we just define <math>SuperLim(a_k)=0</math> for all sequences the 3 properties seem to hold, maybe that <math>SuperLim(a_k)=Lim(a_k)</math> if the sequence does converge should be added? Or maybe just ruling out constant functions is suffient... -[[User:Johnfleming|John]]
** Thanks! You are absolutely right. The question has been modified. [[User:Drorbn|Drorbn]] 15:23, 13 November 2010 (EST)

----

Hi Dror, I understand you want to make it challenging but I think it is also a bit unfair for some students like me. I never learned super-limit function. It is not done in your lecture or discussed in the book. So solving this question totally relies on other background knowledge. I don't have such strong background compared to some other students and I didn't even have a serious lecture on cardinality. I am just hoping since you are asking these kind of questions could you possibly talk about it in class? Trust me Dror it is not that I am not working hard or not trying to struggle but my background is not even enough for me to understand the question. For example I don't know what super-limit function is. I couldn't find it in Munkres book I couldn't find it by googling? Are you talking about limsup?

And what is the generalized definition of limit? I cannot find the definition anywhere.[[User:Xwbdsb|Xwbdsb]] 10:01, 14 November 2010 (EST)

* In the question about super-limits you are asked to prove that "super-limits", objects with the given properties, do not exist. It is not surprising they are not in the literature. Cardinalities are covered in MAT246, which is a pre-requisite to this class, I believe. I'll say more about generalized limits in the coming class tomorrow. [[User:Drorbn|Drorbn]] 10:37, 14 November 2010 (EST)
** Munkres Chapter 1 is all about set theory. In particular sections 6,7 and 9 have discussions about cardinality, this may be a good resource if you are having difficulties with the ideas. - [[User:Johnfleming|John]]

----

*A quick question (and I hope it isn't stupid...), but I'm not sure what "A" is when you talk about the cube <math>I^A</math>. What product of the unit interval are you talking about? --[[User:Wesley|Wesley]] 12:15, 14 November 2010 (EST)
** <math>A</math> is any set whatsoever. So the "cube" <math>I^A</math> is finite dimensional if <math>A</math> is finite, but it is infinite dimensional if <math>A</math> is infinite. [[User:Drorbn|Drorbn]] 13:28, 14 November 2010 (EST)

*Is it true that for question 6 the function f has to be bounded? Since even though b is a bounded sequence but f(b_k) might not be bounded so that the generalized limit is not defined. given that f is continuous. suppose f=1/x and b_k = 1/k. [[User:Xwbdsb|Xwbdsb]] 22:12, 15 November 2010 (EST)
** The function in the example you gave is not continuous; it not even defined at <math>x=0</math>. [[User:Drorbn|Drorbn]] 07:10, 16 November 2010 (EST)

* In problem 1 part c, the errata to Munkres says that should be Hausdorff.
** Thanks! [[User:Drorbn|Drorbn]] 07:10, 16 November 2010 (EST)

**Does the set of all bounded sequences has the same cardinality as the reals?[[User:Xwbdsb|Xwbdsb]] 15:00, 16 November 2010 (EST)

----
Question About Problem 2

Should we assume that {0,1} has the discrete topology and that {0,1}^N has the corresponding product topology and should we also assume that C has the subspace topology that it inherits from R? Thank you.

* Yes and yes. [[User:Drorbn|Drorbn]] 06:59, 17 November 2010 (EST)

Question About Problem 3

If F if the extension of the function f to (beta)N, then do we need to ensure that if mu is in (beta)N \ N, then F(mu) is in [0,1]^A ? (This is similar to how in class we extended a bounded function b to b_tilda but then we did not guarantee that b_tilda was bounded so then the range for b_tilda can be R but may not be any closed interval in R.) Thank you.

<math>\tilde b </math> has to be bounded because <math>\beta{\mathbb N}</math> is compact. [[User:Ian|Ian]] 18:54, 17 November 2010 (EST)

10-327/Homework Assignment 4

2010-10-27T23:49:35Z

Ian: /* Questions */

{{10-327/Navigation}}

===Reading===
Read sections 23 through 25 in Munkres' textbook (Topology, 2nd edition). Remember that reading math isn't like reading a novel! If you read a novel and miss a few details most likely you'll still understand the novel. But if you miss a few details in a math text, often you'll miss everything that follows. So reading math takes reading and rereading and rerereading and a lot of thought about what you've read. Also, '''preread''' sections 26 through 27, just to get a feel for the future.

===Doing===
Solve and submit problems 1-3 and 8-10 Munkres' book, pages 157-158.

===Due date===
This assignment is due at the end of class on Monday, October 25, 2010.

===Suggestions for Good Deeds===
Annotate our Monday videos (starting with {{10-327/vp|0927}}) in a manner similar to (say) {{dbnvp link|AKT-090910-1|AKT-090910-1}}, and/or add links to the blackboard shots, in a manner similar to {{dbnvp link|Alekseev-1006-1|Alekseev-1006-1}}. Also, make ''constructive'' suggestions to me, {{Dror}} and / or the videographer, Qian (Sindy) Li, on how to improve the videos and / or the software used to display them. Note that "constructive" means also, "something that can be implemented relatively easily in the real world, given limited resources".

{{Template:10-327:Dror/Students Divider}}

===Questions===
# Hi, quick question. I am wondering if the term test will cover the material on this assignment, or only the material before the assignment. Thanks! Jason.
#* The term test will cover everything including Monday October 25 and including this assignment. [[User:Drorbn|Drorbn]] 13:42, 23 October 2010 (EDT)
# In EXAMPLE 7 on page 151 Munkres claims that Rn~ is ['clearly' :)] homeomorphic to Rn: where Rn~ consists of all sequences x=(x1,x2,x3,...) with xi=0 for i>n, and Rn consists of all sequences x=(x1,x2,...xn). Why are they homeomorphic ?? Thank you kindly. Oliviu.

RE: 2) Let <math>F :\tilde R^n \rightarrow R^n</math> be defined as <math>F(x)= \prod_{i=1}^{n} \pi_i (x)</math> and let <math>F^{-1} : R^n \rightarrow \tilde R^n</math> be defined as <math>F^{-1}(x)= \prod_{i \in Z_+} f_i (x)</math> where <math> f_i (x) = \pi_i (x) </math> if <math> 1 \le i \le n </math> and <math> f_i(x)=0 </math> otherwise. Then both <math> F </math> and <math> F^{-1} </math> are continuous, because we are working in the product topology and the component functions, namely the projection function and constant function are continuous. Also <math> F </math> is a bijection because <math> F(F^{-1}(x_1, \ldots, x_n))=(x_1, \ldots, x_n) </math> and <math> F^{-1}(F(x_1, \ldots, x_n, 0,0, \ldots))=(x_1, \ldots, x_n, 0,0, \ldots) </math>, i.e <math> F </math> has a left and right inverse. So <math> F </math> is a homeomorphism between the two spaces. Quick question is there a nicer way of writing math than using the math tag? [[User:Ian|Ian]] 16:03, 22 October 2010 (EDT)
**Hi Ian I don't believe what you have said.</math>\tilde R^n</math> has the subspace topology from </math>R^\omega</math> where we indeed put product topology on it. This is not the same as </math>\tilde R^n</math> has the product topology. -Kai
***Kai I don't understand what you mean by not the same can you clarify? Do you mean they are not homeomorphic, i.e. there is something wrong with the functions I provided? [[User:Ian|Ian]] 17:47, 27 October 2010 (EDT)
****Hi Kai I see what you are trying to say now. Theorem 19.3 suggests they are they same. [[User:Ian|Ian]] 19:49, 27 October 2010 (EDT)

3)Question. Suppose we have a function f going from topological space X to Y which is not onto and a function g going from Y to Z. Could I still define the composition of f and g? i.e. g circle f? -Kai [[User:Xwbdsb|Xwbdsb]] 19:19, 22 October 2010 (EDT)
*If I understand your question, I don't see why not...think about <math>\mathbb{R}</math> for example. <math>f(x)=x^2</math> is not onto, then let <math>g(x)=e^x</math> then g compose f is <math>e^{x^2}</math> - John
**I agree but look at munkre's page 17 last sentence. Note that g compose with f is defined only when the range of f equals the domain of g. So I just want to confirm with Dror if there is something wrong here.
***Touche, I see your point...that is strange - John

4)Question about the proof for [0,1] being connected. A few details are omitted. why would a closed subset of [0,1] contain its supremum? Also why [0,g_0] being a subset of A follows automatically after we showed that g_0 is in A? -Kai
*1. Suppose <math>S </math> is closed in <math> [0,1]. \Rightarrow S^C </math> is open. If <math> sup(S)=\alpha \notin S \Rightarrow \exists r>0 </math> s.t. <math> B(\alpha, r) \subset S^C \Rightarrow \alpha - 0.5r \in S^C \Rightarrow \alpha - 0.5r < \alpha </math> is an upper bound for S. <math> \Rightarrow \Leftarrow </math>
*2. Recall that <math> G = \{g | [0,g] \subset A\}; g_0 = sup(G) \Rightarrow \forall g < g_0, [0,g] \subset A \Rightarrow [0, g_0) \subset A </math>. So, if <math> g_0 \in A \Rightarrow [0,g_0] \subset A</math>. -Frank [[User:Fzhao|Fzhao]] 23:50, 22 October 2010 (EDT)
**Thanks Frank. But I don't think your solution is convincing enough. \alpha - 0.5r is indeed not in S but why can you say it is an upper bound for S? Remember S could be rather complicated set all you know is that it is closed.
for 2 why is \Rightarrow \forall g < g_0, [0,g]? even if g_0 is sup(G) that does not mean anything less than g_0 would be in G. Consider [0,1] union {3}.

*Well, for the first question, not only is <math> \alpha - 0.5r </math> not in S, but neither is anything in <math> B(\alpha,r) </math>, since <math> S^C </math> is open. There can be no elements <math> \geq \alpha </math> in S because it's the supremum. Recall also we're working in the Reals. <math> \Rightarrow [\alpha - 0.5r, \infty) \subset S^C. </math>

*For the second question, notice that the supremum is the least upper bound (of G), so <math> \forall r > 0, \exists g \in G \cap B(g_0,r) \Rightarrow [0,g] \subset A \Rightarrow [0,g'] \subset A, \forall g' < g \Rightarrow g' \in G \Rightarrow \forall g < g_0 </math>, take <math> r = 0.5(g_0 - g) \Rightarrow \exists a \in (g_0 - r, g_0) \cap G, \Rightarrow [0,g] \subset [0,a] \subset A.</math>
Specifically, in your counterexample, if 3 is in G, then anything less than 3 is also in G by construction of G.

*Perhaps (I'm guessing here) you might have found supremum to be a confusing notion. If this is the case, have no fear, there's a chapter on supremum in Spivak's book Calculus. You can probably find one in the math library. - Frank [[User:Fzhao|Fzhao]] 09:58, 23 October 2010 (EDT)

*Question 1(b) on HW4. x stands for a point in S^1 but what does it mean by -x? -Kai
** <math>S^1\subset{\mathbb R}^2</math>, and <math>{\mathbb R}^2</math> is a vector space. [[User:Drorbn|Drorbn]] 13:42, 23 October 2010 (EDT)

*Question. During the proof of A \subset B \subset of A closure when they are all subspaces of X. If A is connected and so is B. I think there is loss of generality. We should prove any non-empty clopen set is B. But we are only proving those clopen sets whose intersection with A is non-empty are B. That is not enough because we could also have non-empty clopen set in B which does not intersect A. -Kai
** If <math>B</math> doesn't intersect <math>A</math> it's complement will. [[User:Drorbn|Drorbn]] 13:42, 23 October 2010 (EDT)

*Question. Consider the usual basis for infinite product topology that we put on an infinite cartesian product of topological spaces. The basis is just the topology generated minus the empty set right? [[User:Xwbdsb|Xwbdsb]] 23:23, 23 October 2010 (EDT)
** No. In an infinite (or even finite) product there are open sets which are unions of cylinders but are not themselves cylinders. A ball in the plane is a union of boxes but it is not a box. Another example is <math>{\mathbb R}^\omega-\{\bar{0}\}</math>. [[User:Drorbn|Drorbn]] 11:45, 24 October 2010 (EDT)

* Another question. We proved before if we have a collection of metrizable topological spaces then if we put cylinder topology on the infinite cartesian product then the resulting topological space is going to be metrizable. Then why isnt R^R with cylinder topology metrizable a contradiction since each copy of R is metrizable and R^R is just the infinite cartesian product of copies of R with indexing set R. -Kai[[User:Xwbdsb|Xwbdsb]] 22:58, 24 October 2010 (EDT)
** The theorem in class was about ''countable'' products of metric spaces. [[User:Drorbn|Drorbn]] 10:00, 25 October 2010 (EDT)
***For the first half of the proof we don't require countability assumption right because if the product space is metrizable even if it is uncountable product we can still create a homeomorphic subspace to each of the individual topological spaces? -Kai [[User:Xwbdsb|Xwbdsb]] 10:37, 25 October 2010 (EDT)

* Is it true that if one general topological space is metrizable and if one general topological space is homeomorphic to it then it is metrizable? Why? Metrizability is not just topological property? It involves the definition of metric which depends on the structure of a set.?? -Kai [[User:Xwbdsb|Xwbdsb]] 00:25, 25 October 2010 (EDT)
** "Metrizable" is a topological property, it means "having some metric which induces the topology". If <math>X</math> is metrizable and <math>Y</math> is homeomorphic to it, a metric on <math>X</math> composed with the homeomorphism from <math>Y</math> to <math>X</math> defines a metric on <math>Y</math> which induces the topology of <math>Y</math>. [[User:Drorbn|Drorbn]] 10:00, 25 October 2010 (EDT)

*Question. When proving if X is a metrizable space and A is a countably infinite set in X. The cardinality of the sequential closure of A is less or equal to the cardinality of R we used the fact that a metrizable space is Hausdorff right? Or you can say that the number of limit points should be less or equal to the number of converging sequences... -Kai
** Yes we use the fact that metric spaces are <math>T_2</math>. [[User:Drorbn|Drorbn]] 20:30, 26 October 2010 (EDT)

*Question. When proving topologist's sine curve is not path connected in page 157. Are we using the fact that for a cts function in Euclidean space if x_n converges to x then f(x_n) has to converge to f(x)? i.e. any open nbd of f(x) contains all but finitely many terms of f(x_n)? -Kai[[User:Xwbdsb|Xwbdsb]] 14:01, 27 October 2010 (EDT)
** We did not complete the proof of that fact in class. There are many ways to complete it, some use what you wrote and some don't. [[User:Drorbn|Drorbn]] 14:48, 27 October 2010 (EDT)
*Question. Is f(t) = (t,t,t...) cts in the R^\omega with uniform topology?.... -Kai [[User:Xwbdsb|Xwbdsb]] 15:23, 27 October 2010 (EDT)

10-327/Homework Assignment 4

2010-10-27T23:48:31Z

Ian: /* Questions */

{{10-327/Navigation}}

===Reading===
Read sections 23 through 25 in Munkres' textbook (Topology, 2nd edition). Remember that reading math isn't like reading a novel! If you read a novel and miss a few details most likely you'll still understand the novel. But if you miss a few details in a math text, often you'll miss everything that follows. So reading math takes reading and rereading and rerereading and a lot of thought about what you've read. Also, '''preread''' sections 26 through 27, just to get a feel for the future.

===Doing===
Solve and submit problems 1-3 and 8-10 Munkres' book, pages 157-158.

===Due date===
This assignment is due at the end of class on Monday, October 25, 2010.

===Suggestions for Good Deeds===
Annotate our Monday videos (starting with {{10-327/vp|0927}}) in a manner similar to (say) {{dbnvp link|AKT-090910-1|AKT-090910-1}}, and/or add links to the blackboard shots, in a manner similar to {{dbnvp link|Alekseev-1006-1|Alekseev-1006-1}}. Also, make ''constructive'' suggestions to me, {{Dror}} and / or the videographer, Qian (Sindy) Li, on how to improve the videos and / or the software used to display them. Note that "constructive" means also, "something that can be implemented relatively easily in the real world, given limited resources".

{{Template:10-327:Dror/Students Divider}}

===Questions===
# Hi, quick question. I am wondering if the term test will cover the material on this assignment, or only the material before the assignment. Thanks! Jason.
#* The term test will cover everything including Monday October 25 and including this assignment. [[User:Drorbn|Drorbn]] 13:42, 23 October 2010 (EDT)
# In EXAMPLE 7 on page 151 Munkres claims that Rn~ is ['clearly' :)] homeomorphic to Rn: where Rn~ consists of all sequences x=(x1,x2,x3,...) with xi=0 for i>n, and Rn consists of all sequences x=(x1,x2,...xn). Why are they homeomorphic ?? Thank you kindly. Oliviu.

RE: 2) Let <math>F :\tilde R^n \rightarrow R^n</math> be defined as <math>F(x)= \prod_{i=1}^{n} \pi_i (x)</math> and let <math>F^{-1} : R^n \rightarrow \tilde R^n</math> be defined as <math>F^{-1}(x)= \prod_{i \in Z_+} f_i (x)</math> where <math> f_i (x) = \pi_i (x) </math> if <math> 1 \le i \le n </math> and <math> f_i(x)=0 </math> otherwise. Then both <math> F </math> and <math> F^{-1} </math> are continuous, because we are working in the product topology and the component functions, namely the projection function and constant function are continuous. Also <math> F </math> is a bijection because <math> F(F^{-1}(x_1, \ldots, x_n))=(x_1, \ldots, x_n) </math> and <math> F^{-1}(F(x_1, \ldots, x_n, 0,0, \ldots))=(x_1, \ldots, x_n, 0,0, \ldots) </math>, i.e <math> F </math> has a left and right inverse. So <math> F </math> is a homeomorphism between the two spaces. Quick question is there a nicer way of writing math than using the math tag? [[User:Ian|Ian]] 16:03, 22 October 2010 (EDT)
**Hi Ian I don't believe what you have said.</math>\tilde R^n</math> has the subspace topology from </math>R^\omega</math> where we indeed put product topology on it. This is not the same as </math>\tilde R^n</math> has the product topology. -Kai
***Kai I don't understand what you mean by not the same can you clarify? Do you mean they are not homeomorphic, i.e. there is something wrong with the functions I provided? [[User:Ian|Ian]] 17:47, 27 October 2010 (EDT)
****Hi Kai I see what you are trying to say now. Theorem 19.3 suggests they are they same.

3)Question. Suppose we have a function f going from topological space X to Y which is not onto and a function g going from Y to Z. Could I still define the composition of f and g? i.e. g circle f? -Kai [[User:Xwbdsb|Xwbdsb]] 19:19, 22 October 2010 (EDT)
*If I understand your question, I don't see why not...think about <math>\mathbb{R}</math> for example. <math>f(x)=x^2</math> is not onto, then let <math>g(x)=e^x</math> then g compose f is <math>e^{x^2}</math> - John
**I agree but look at munkre's page 17 last sentence. Note that g compose with f is defined only when the range of f equals the domain of g. So I just want to confirm with Dror if there is something wrong here.
***Touche, I see your point...that is strange - John

4)Question about the proof for [0,1] being connected. A few details are omitted. why would a closed subset of [0,1] contain its supremum? Also why [0,g_0] being a subset of A follows automatically after we showed that g_0 is in A? -Kai
*1. Suppose <math>S </math> is closed in <math> [0,1]. \Rightarrow S^C </math> is open. If <math> sup(S)=\alpha \notin S \Rightarrow \exists r>0 </math> s.t. <math> B(\alpha, r) \subset S^C \Rightarrow \alpha - 0.5r \in S^C \Rightarrow \alpha - 0.5r < \alpha </math> is an upper bound for S. <math> \Rightarrow \Leftarrow </math>
*2. Recall that <math> G = \{g | [0,g] \subset A\}; g_0 = sup(G) \Rightarrow \forall g < g_0, [0,g] \subset A \Rightarrow [0, g_0) \subset A </math>. So, if <math> g_0 \in A \Rightarrow [0,g_0] \subset A</math>. -Frank [[User:Fzhao|Fzhao]] 23:50, 22 October 2010 (EDT)
**Thanks Frank. But I don't think your solution is convincing enough. \alpha - 0.5r is indeed not in S but why can you say it is an upper bound for S? Remember S could be rather complicated set all you know is that it is closed.
for 2 why is \Rightarrow \forall g < g_0, [0,g]? even if g_0 is sup(G) that does not mean anything less than g_0 would be in G. Consider [0,1] union {3}.

*Well, for the first question, not only is <math> \alpha - 0.5r </math> not in S, but neither is anything in <math> B(\alpha,r) </math>, since <math> S^C </math> is open. There can be no elements <math> \geq \alpha </math> in S because it's the supremum. Recall also we're working in the Reals. <math> \Rightarrow [\alpha - 0.5r, \infty) \subset S^C. </math>

*For the second question, notice that the supremum is the least upper bound (of G), so <math> \forall r > 0, \exists g \in G \cap B(g_0,r) \Rightarrow [0,g] \subset A \Rightarrow [0,g'] \subset A, \forall g' < g \Rightarrow g' \in G \Rightarrow \forall g < g_0 </math>, take <math> r = 0.5(g_0 - g) \Rightarrow \exists a \in (g_0 - r, g_0) \cap G, \Rightarrow [0,g] \subset [0,a] \subset A.</math>
Specifically, in your counterexample, if 3 is in G, then anything less than 3 is also in G by construction of G.

*Perhaps (I'm guessing here) you might have found supremum to be a confusing notion. If this is the case, have no fear, there's a chapter on supremum in Spivak's book Calculus. You can probably find one in the math library. - Frank [[User:Fzhao|Fzhao]] 09:58, 23 October 2010 (EDT)

*Question 1(b) on HW4. x stands for a point in S^1 but what does it mean by -x? -Kai
** <math>S^1\subset{\mathbb R}^2</math>, and <math>{\mathbb R}^2</math> is a vector space. [[User:Drorbn|Drorbn]] 13:42, 23 October 2010 (EDT)

*Question. During the proof of A \subset B \subset of A closure when they are all subspaces of X. If A is connected and so is B. I think there is loss of generality. We should prove any non-empty clopen set is B. But we are only proving those clopen sets whose intersection with A is non-empty are B. That is not enough because we could also have non-empty clopen set in B which does not intersect A. -Kai
** If <math>B</math> doesn't intersect <math>A</math> it's complement will. [[User:Drorbn|Drorbn]] 13:42, 23 October 2010 (EDT)

*Question. Consider the usual basis for infinite product topology that we put on an infinite cartesian product of topological spaces. The basis is just the topology generated minus the empty set right? [[User:Xwbdsb|Xwbdsb]] 23:23, 23 October 2010 (EDT)
** No. In an infinite (or even finite) product there are open sets which are unions of cylinders but are not themselves cylinders. A ball in the plane is a union of boxes but it is not a box. Another example is <math>{\mathbb R}^\omega-\{\bar{0}\}</math>. [[User:Drorbn|Drorbn]] 11:45, 24 October 2010 (EDT)

* Another question. We proved before if we have a collection of metrizable topological spaces then if we put cylinder topology on the infinite cartesian product then the resulting topological space is going to be metrizable. Then why isnt R^R with cylinder topology metrizable a contradiction since each copy of R is metrizable and R^R is just the infinite cartesian product of copies of R with indexing set R. -Kai[[User:Xwbdsb|Xwbdsb]] 22:58, 24 October 2010 (EDT)
** The theorem in class was about ''countable'' products of metric spaces. [[User:Drorbn|Drorbn]] 10:00, 25 October 2010 (EDT)
***For the first half of the proof we don't require countability assumption right because if the product space is metrizable even if it is uncountable product we can still create a homeomorphic subspace to each of the individual topological spaces? -Kai [[User:Xwbdsb|Xwbdsb]] 10:37, 25 October 2010 (EDT)

* Is it true that if one general topological space is metrizable and if one general topological space is homeomorphic to it then it is metrizable? Why? Metrizability is not just topological property? It involves the definition of metric which depends on the structure of a set.?? -Kai [[User:Xwbdsb|Xwbdsb]] 00:25, 25 October 2010 (EDT)
** "Metrizable" is a topological property, it means "having some metric which induces the topology". If <math>X</math> is metrizable and <math>Y</math> is homeomorphic to it, a metric on <math>X</math> composed with the homeomorphism from <math>Y</math> to <math>X</math> defines a metric on <math>Y</math> which induces the topology of <math>Y</math>. [[User:Drorbn|Drorbn]] 10:00, 25 October 2010 (EDT)

*Question. When proving if X is a metrizable space and A is a countably infinite set in X. The cardinality of the sequential closure of A is less or equal to the cardinality of R we used the fact that a metrizable space is Hausdorff right? Or you can say that the number of limit points should be less or equal to the number of converging sequences... -Kai
** Yes we use the fact that metric spaces are <math>T_2</math>. [[User:Drorbn|Drorbn]] 20:30, 26 October 2010 (EDT)

*Question. When proving topologist's sine curve is not path connected in page 157. Are we using the fact that for a cts function in Euclidean space if x_n converges to x then f(x_n) has to converge to f(x)? i.e. any open nbd of f(x) contains all but finitely many terms of f(x_n)? -Kai[[User:Xwbdsb|Xwbdsb]] 14:01, 27 October 2010 (EDT)
** We did not complete the proof of that fact in class. There are many ways to complete it, some use what you wrote and some don't. [[User:Drorbn|Drorbn]] 14:48, 27 October 2010 (EDT)
*Question. Is f(t) = (t,t,t...) cts in the R^\omega with uniform topology?.... -Kai [[User:Xwbdsb|Xwbdsb]] 15:23, 27 October 2010 (EDT)

10-327/Homework Assignment 4

2010-10-27T21:47:58Z

Ian: /* Questions */

{{10-327/Navigation}}

===Reading===
Read sections 23 through 25 in Munkres' textbook (Topology, 2nd edition). Remember that reading math isn't like reading a novel! If you read a novel and miss a few details most likely you'll still understand the novel. But if you miss a few details in a math text, often you'll miss everything that follows. So reading math takes reading and rereading and rerereading and a lot of thought about what you've read. Also, '''preread''' sections 26 through 27, just to get a feel for the future.

===Doing===
Solve and submit problems 1-3 and 8-10 Munkres' book, pages 157-158.

===Due date===
This assignment is due at the end of class on Monday, October 25, 2010.

===Suggestions for Good Deeds===
Annotate our Monday videos (starting with {{10-327/vp|0927}}) in a manner similar to (say) {{dbnvp link|AKT-090910-1|AKT-090910-1}}, and/or add links to the blackboard shots, in a manner similar to {{dbnvp link|Alekseev-1006-1|Alekseev-1006-1}}. Also, make ''constructive'' suggestions to me, {{Dror}} and / or the videographer, Qian (Sindy) Li, on how to improve the videos and / or the software used to display them. Note that "constructive" means also, "something that can be implemented relatively easily in the real world, given limited resources".

{{Template:10-327:Dror/Students Divider}}

===Questions===
# Hi, quick question. I am wondering if the term test will cover the material on this assignment, or only the material before the assignment. Thanks! Jason.
#* The term test will cover everything including Monday October 25 and including this assignment. [[User:Drorbn|Drorbn]] 13:42, 23 October 2010 (EDT)
# In EXAMPLE 7 on page 151 Munkres claims that Rn~ is ['clearly' :)] homeomorphic to Rn: where Rn~ consists of all sequences x=(x1,x2,x3,...) with xi=0 for i>n, and Rn consists of all sequences x=(x1,x2,...xn). Why are they homeomorphic ?? Thank you kindly. Oliviu.

RE: 2) Let <math>F :\tilde R^n \rightarrow R^n</math> be defined as <math>F(x)= \prod_{i=1}^{n} \pi_i (x)</math> and let <math>F^{-1} : R^n \rightarrow \tilde R^n</math> be defined as <math>F^{-1}(x)= \prod_{i \in Z_+} f_i (x)</math> where <math> f_i (x) = \pi_i (x) </math> if <math> 1 \le i \le n </math> and <math> f_i(x)=0 </math> otherwise. Then both <math> F </math> and <math> F^{-1} </math> are continuous, because we are working in the product topology and the component functions, namely the projection function and constant function are continuous. Also <math> F </math> is a bijection because <math> F(F^{-1}(x_1, \ldots, x_n))=(x_1, \ldots, x_n) </math> and <math> F^{-1}(F(x_1, \ldots, x_n, 0,0, \ldots))=(x_1, \ldots, x_n, 0,0, \ldots) </math>, i.e <math> F </math> has a left and right inverse. So <math> F </math> is a homeomorphism between the two spaces. Quick question is there a nicer way of writing math than using the math tag? [[User:Ian|Ian]] 16:03, 22 October 2010 (EDT)
**Hi Ian I don't believe what you have said.</math>\tilde R^n</math> has the subspace topology from </math>R^\omega</math> where we indeed put product topology on it. This is not the same as </math>\tilde R^n</math> has the product topology. -Kai
***Kai I don't understand what you mean by not the same can you clarify? Do you mean they are not homeomorphic, i.e. there is something wrong with the functions I provided? [[User:Ian|Ian]] 17:47, 27 October 2010 (EDT)

3)Question. Suppose we have a function f going from topological space X to Y which is not onto and a function g going from Y to Z. Could I still define the composition of f and g? i.e. g circle f? -Kai [[User:Xwbdsb|Xwbdsb]] 19:19, 22 October 2010 (EDT)
*If I understand your question, I don't see why not...think about <math>\mathbb{R}</math> for example. <math>f(x)=x^2</math> is not onto, then let <math>g(x)=e^x</math> then g compose f is <math>e^{x^2}</math> - John
**I agree but look at munkre's page 17 last sentence. Note that g compose with f is defined only when the range of f equals the domain of g. So I just want to confirm with Dror if there is something wrong here.
***Touche, I see your point...that is strange - John

4)Question about the proof for [0,1] being connected. A few details are omitted. why would a closed subset of [0,1] contain its supremum? Also why [0,g_0] being a subset of A follows automatically after we showed that g_0 is in A? -Kai
*1. Suppose <math>S </math> is closed in <math> [0,1]. \Rightarrow S^C </math> is open. If <math> sup(S)=\alpha \notin S \Rightarrow \exists r>0 </math> s.t. <math> B(\alpha, r) \subset S^C \Rightarrow \alpha - 0.5r \in S^C \Rightarrow \alpha - 0.5r < \alpha </math> is an upper bound for S. <math> \Rightarrow \Leftarrow </math>
*2. Recall that <math> G = \{g | [0,g] \subset A\}; g_0 = sup(G) \Rightarrow \forall g < g_0, [0,g] \subset A \Rightarrow [0, g_0) \subset A </math>. So, if <math> g_0 \in A \Rightarrow [0,g_0] \subset A</math>. -Frank [[User:Fzhao|Fzhao]] 23:50, 22 October 2010 (EDT)
**Thanks Frank. But I don't think your solution is convincing enough. \alpha - 0.5r is indeed not in S but why can you say it is an upper bound for S? Remember S could be rather complicated set all you know is that it is closed.
for 2 why is \Rightarrow \forall g < g_0, [0,g]? even if g_0 is sup(G) that does not mean anything less than g_0 would be in G. Consider [0,1] union {3}.

*Well, for the first question, not only is <math> \alpha - 0.5r </math> not in S, but neither is anything in <math> B(\alpha,r) </math>, since <math> S^C </math> is open. There can be no elements <math> \geq \alpha </math> in S because it's the supremum. Recall also we're working in the Reals. <math> \Rightarrow [\alpha - 0.5r, \infty) \subset S^C. </math>

*For the second question, notice that the supremum is the least upper bound (of G), so <math> \forall r > 0, \exists g \in G \cap B(g_0,r) \Rightarrow [0,g] \subset A \Rightarrow [0,g'] \subset A, \forall g' < g \Rightarrow g' \in G \Rightarrow \forall g < g_0 </math>, take <math> r = 0.5(g_0 - g) \Rightarrow \exists a \in (g_0 - r, g_0) \cap G, \Rightarrow [0,g] \subset [0,a] \subset A.</math>
Specifically, in your counterexample, if 3 is in G, then anything less than 3 is also in G by construction of G.

*Perhaps (I'm guessing here) you might have found supremum to be a confusing notion. If this is the case, have no fear, there's a chapter on supremum in Spivak's book Calculus. You can probably find one in the math library. - Frank [[User:Fzhao|Fzhao]] 09:58, 23 October 2010 (EDT)

*Question 1(b) on HW4. x stands for a point in S^1 but what does it mean by -x? -Kai
** <math>S^1\subset{\mathbb R}^2</math>, and <math>{\mathbb R}^2</math> is a vector space. [[User:Drorbn|Drorbn]] 13:42, 23 October 2010 (EDT)

*Question. During the proof of A \subset B \subset of A closure when they are all subspaces of X. If A is connected and so is B. I think there is loss of generality. We should prove any non-empty clopen set is B. But we are only proving those clopen sets whose intersection with A is non-empty are B. That is not enough because we could also have non-empty clopen set in B which does not intersect A. -Kai
** If <math>B</math> doesn't intersect <math>A</math> it's complement will. [[User:Drorbn|Drorbn]] 13:42, 23 October 2010 (EDT)

*Question. Consider the usual basis for infinite product topology that we put on an infinite cartesian product of topological spaces. The basis is just the topology generated minus the empty set right? [[User:Xwbdsb|Xwbdsb]] 23:23, 23 October 2010 (EDT)
** No. In an infinite (or even finite) product there are open sets which are unions of cylinders but are not themselves cylinders. A ball in the plane is a union of boxes but it is not a box. Another example is <math>{\mathbb R}^\omega-\{\bar{0}\}</math>. [[User:Drorbn|Drorbn]] 11:45, 24 October 2010 (EDT)

* Another question. We proved before if we have a collection of metrizable topological spaces then if we put cylinder topology on the infinite cartesian product then the resulting topological space is going to be metrizable. Then why isnt R^R with cylinder topology metrizable a contradiction since each copy of R is metrizable and R^R is just the infinite cartesian product of copies of R with indexing set R. -Kai[[User:Xwbdsb|Xwbdsb]] 22:58, 24 October 2010 (EDT)
** The theorem in class was about ''countable'' products of metric spaces. [[User:Drorbn|Drorbn]] 10:00, 25 October 2010 (EDT)
***For the first half of the proof we don't require countability assumption right because if the product space is metrizable even if it is uncountable product we can still create a homeomorphic subspace to each of the individual topological spaces? -Kai [[User:Xwbdsb|Xwbdsb]] 10:37, 25 October 2010 (EDT)

* Is it true that if one general topological space is metrizable and if one general topological space is homeomorphic to it then it is metrizable? Why? Metrizability is not just topological property? It involves the definition of metric which depends on the structure of a set.?? -Kai [[User:Xwbdsb|Xwbdsb]] 00:25, 25 October 2010 (EDT)
** "Metrizable" is a topological property, it means "having some metric which induces the topology". If <math>X</math> is metrizable and <math>Y</math> is homeomorphic to it, a metric on <math>X</math> composed with the homeomorphism from <math>Y</math> to <math>X</math> defines a metric on <math>Y</math> which induces the topology of <math>Y</math>. [[User:Drorbn|Drorbn]] 10:00, 25 October 2010 (EDT)

*Question. When proving if X is a metrizable space and A is a countably infinite set in X. The cardinality of the sequential closure of A is less or equal to the cardinality of R we used the fact that a metrizable space is Hausdorff right? Or you can say that the number of limit points should be less or equal to the number of converging sequences... -Kai
** Yes we use the fact that metric spaces are <math>T_2</math>. [[User:Drorbn|Drorbn]] 20:30, 26 October 2010 (EDT)

*Question. When proving topologist's sine curve is not path connected in page 157. Are we using the fact that for a cts function in Euclidean space if x_n converges to x then f(x_n) has to converge to f(x)? i.e. any open nbd of f(x) contains all but finitely many terms of f(x_n)? -Kai[[User:Xwbdsb|Xwbdsb]] 14:01, 27 October 2010 (EDT)
** We did not complete the proof of that fact in class. There are many ways to complete it, some use what you wrote and some don't. [[User:Drorbn|Drorbn]] 14:48, 27 October 2010 (EDT)
*Question. Is f(t) = (t,t,t...) cts in the R^\omega with uniform topology?.... -Kai [[User:Xwbdsb|Xwbdsb]] 15:23, 27 October 2010 (EDT)

10-327/Homework Assignment 4

2010-10-27T21:47:37Z

Ian: /* Questions */

{{10-327/Navigation}}

===Reading===
Read sections 23 through 25 in Munkres' textbook (Topology, 2nd edition). Remember that reading math isn't like reading a novel! If you read a novel and miss a few details most likely you'll still understand the novel. But if you miss a few details in a math text, often you'll miss everything that follows. So reading math takes reading and rereading and rerereading and a lot of thought about what you've read. Also, '''preread''' sections 26 through 27, just to get a feel for the future.

===Doing===
Solve and submit problems 1-3 and 8-10 Munkres' book, pages 157-158.

===Due date===
This assignment is due at the end of class on Monday, October 25, 2010.

===Suggestions for Good Deeds===
Annotate our Monday videos (starting with {{10-327/vp|0927}}) in a manner similar to (say) {{dbnvp link|AKT-090910-1|AKT-090910-1}}, and/or add links to the blackboard shots, in a manner similar to {{dbnvp link|Alekseev-1006-1|Alekseev-1006-1}}. Also, make ''constructive'' suggestions to me, {{Dror}} and / or the videographer, Qian (Sindy) Li, on how to improve the videos and / or the software used to display them. Note that "constructive" means also, "something that can be implemented relatively easily in the real world, given limited resources".

{{Template:10-327:Dror/Students Divider}}

===Questions===
# Hi, quick question. I am wondering if the term test will cover the material on this assignment, or only the material before the assignment. Thanks! Jason.
#* The term test will cover everything including Monday October 25 and including this assignment. [[User:Drorbn|Drorbn]] 13:42, 23 October 2010 (EDT)
# In EXAMPLE 7 on page 151 Munkres claims that Rn~ is ['clearly' :)] homeomorphic to Rn: where Rn~ consists of all sequences x=(x1,x2,x3,...) with xi=0 for i>n, and Rn consists of all sequences x=(x1,x2,...xn). Why are they homeomorphic ?? Thank you kindly. Oliviu.

RE: 2) Let <math>F :\tilde R^n \rightarrow R^n</math> be defined as <math>F(x)= \prod_{i=1}^{n} \pi_i (x)</math> and let <math>F^{-1} : R^n \rightarrow \tilde R^n</math> be defined as <math>F^{-1}(x)= \prod_{i \in Z_+} f_i (x)</math> where <math> f_i (x) = \pi_i (x) </math> if <math> 1 \le i \le n </math> and <math> f_i(x)=0 </math> otherwise. Then both <math> F </math> and <math> F^{-1} </math> are continuous, because we are working in the product topology and the component functions, namely the projection function and constant function are continuous. Also <math> F </math> is a bijection because <math> F(F^{-1}(x_1, \ldots, x_n))=(x_1, \ldots, x_n) </math> and <math> F^{-1}(F(x_1, \ldots, x_n, 0,0, \ldots))=(x_1, \ldots, x_n, 0,0, \ldots) </math>, i.e <math> F </math> has a left and right inverse. So <math> F </math> is a homeomorphism between the two spaces. Quick question is there a nicer way of writing math than using the math tag? [[User:Ian|Ian]] 16:03, 22 October 2010 (EDT)
**Hi Ian I don't believe what you have said.</math>\tilde R^n</math> has the subspace topology from </math>R^\omega</math> where we indeed put product topology on it. This is not the same as </math>\tilde R^n</math> has the product topology. -Kai

***Kai I don't understand what you mean by not the same can you clarify? Do you mean they are not homeomorphic, i.e. there is something wrong with the functions I provided? [[User:Ian|Ian]] 17:47, 27 October 2010 (EDT)

3)Question. Suppose we have a function f going from topological space X to Y which is not onto and a function g going from Y to Z. Could I still define the composition of f and g? i.e. g circle f? -Kai [[User:Xwbdsb|Xwbdsb]] 19:19, 22 October 2010 (EDT)
*If I understand your question, I don't see why not...think about <math>\mathbb{R}</math> for example. <math>f(x)=x^2</math> is not onto, then let <math>g(x)=e^x</math> then g compose f is <math>e^{x^2}</math> - John
**I agree but look at munkre's page 17 last sentence. Note that g compose with f is defined only when the range of f equals the domain of g. So I just want to confirm with Dror if there is something wrong here.
***Touche, I see your point...that is strange - John

4)Question about the proof for [0,1] being connected. A few details are omitted. why would a closed subset of [0,1] contain its supremum? Also why [0,g_0] being a subset of A follows automatically after we showed that g_0 is in A? -Kai
*1. Suppose <math>S </math> is closed in <math> [0,1]. \Rightarrow S^C </math> is open. If <math> sup(S)=\alpha \notin S \Rightarrow \exists r>0 </math> s.t. <math> B(\alpha, r) \subset S^C \Rightarrow \alpha - 0.5r \in S^C \Rightarrow \alpha - 0.5r < \alpha </math> is an upper bound for S. <math> \Rightarrow \Leftarrow </math>
*2. Recall that <math> G = \{g | [0,g] \subset A\}; g_0 = sup(G) \Rightarrow \forall g < g_0, [0,g] \subset A \Rightarrow [0, g_0) \subset A </math>. So, if <math> g_0 \in A \Rightarrow [0,g_0] \subset A</math>. -Frank [[User:Fzhao|Fzhao]] 23:50, 22 October 2010 (EDT)
**Thanks Frank. But I don't think your solution is convincing enough. \alpha - 0.5r is indeed not in S but why can you say it is an upper bound for S? Remember S could be rather complicated set all you know is that it is closed.
for 2 why is \Rightarrow \forall g < g_0, [0,g]? even if g_0 is sup(G) that does not mean anything less than g_0 would be in G. Consider [0,1] union {3}.

*Well, for the first question, not only is <math> \alpha - 0.5r </math> not in S, but neither is anything in <math> B(\alpha,r) </math>, since <math> S^C </math> is open. There can be no elements <math> \geq \alpha </math> in S because it's the supremum. Recall also we're working in the Reals. <math> \Rightarrow [\alpha - 0.5r, \infty) \subset S^C. </math>

*For the second question, notice that the supremum is the least upper bound (of G), so <math> \forall r > 0, \exists g \in G \cap B(g_0,r) \Rightarrow [0,g] \subset A \Rightarrow [0,g'] \subset A, \forall g' < g \Rightarrow g' \in G \Rightarrow \forall g < g_0 </math>, take <math> r = 0.5(g_0 - g) \Rightarrow \exists a \in (g_0 - r, g_0) \cap G, \Rightarrow [0,g] \subset [0,a] \subset A.</math>
Specifically, in your counterexample, if 3 is in G, then anything less than 3 is also in G by construction of G.

*Perhaps (I'm guessing here) you might have found supremum to be a confusing notion. If this is the case, have no fear, there's a chapter on supremum in Spivak's book Calculus. You can probably find one in the math library. - Frank [[User:Fzhao|Fzhao]] 09:58, 23 October 2010 (EDT)

*Question 1(b) on HW4. x stands for a point in S^1 but what does it mean by -x? -Kai
** <math>S^1\subset{\mathbb R}^2</math>, and <math>{\mathbb R}^2</math> is a vector space. [[User:Drorbn|Drorbn]] 13:42, 23 October 2010 (EDT)

*Question. During the proof of A \subset B \subset of A closure when they are all subspaces of X. If A is connected and so is B. I think there is loss of generality. We should prove any non-empty clopen set is B. But we are only proving those clopen sets whose intersection with A is non-empty are B. That is not enough because we could also have non-empty clopen set in B which does not intersect A. -Kai
** If <math>B</math> doesn't intersect <math>A</math> it's complement will. [[User:Drorbn|Drorbn]] 13:42, 23 October 2010 (EDT)

*Question. Consider the usual basis for infinite product topology that we put on an infinite cartesian product of topological spaces. The basis is just the topology generated minus the empty set right? [[User:Xwbdsb|Xwbdsb]] 23:23, 23 October 2010 (EDT)
** No. In an infinite (or even finite) product there are open sets which are unions of cylinders but are not themselves cylinders. A ball in the plane is a union of boxes but it is not a box. Another example is <math>{\mathbb R}^\omega-\{\bar{0}\}</math>. [[User:Drorbn|Drorbn]] 11:45, 24 October 2010 (EDT)

* Another question. We proved before if we have a collection of metrizable topological spaces then if we put cylinder topology on the infinite cartesian product then the resulting topological space is going to be metrizable. Then why isnt R^R with cylinder topology metrizable a contradiction since each copy of R is metrizable and R^R is just the infinite cartesian product of copies of R with indexing set R. -Kai[[User:Xwbdsb|Xwbdsb]] 22:58, 24 October 2010 (EDT)
** The theorem in class was about ''countable'' products of metric spaces. [[User:Drorbn|Drorbn]] 10:00, 25 October 2010 (EDT)
***For the first half of the proof we don't require countability assumption right because if the product space is metrizable even if it is uncountable product we can still create a homeomorphic subspace to each of the individual topological spaces? -Kai [[User:Xwbdsb|Xwbdsb]] 10:37, 25 October 2010 (EDT)

* Is it true that if one general topological space is metrizable and if one general topological space is homeomorphic to it then it is metrizable? Why? Metrizability is not just topological property? It involves the definition of metric which depends on the structure of a set.?? -Kai [[User:Xwbdsb|Xwbdsb]] 00:25, 25 October 2010 (EDT)
** "Metrizable" is a topological property, it means "having some metric which induces the topology". If <math>X</math> is metrizable and <math>Y</math> is homeomorphic to it, a metric on <math>X</math> composed with the homeomorphism from <math>Y</math> to <math>X</math> defines a metric on <math>Y</math> which induces the topology of <math>Y</math>. [[User:Drorbn|Drorbn]] 10:00, 25 October 2010 (EDT)

*Question. When proving if X is a metrizable space and A is a countably infinite set in X. The cardinality of the sequential closure of A is less or equal to the cardinality of R we used the fact that a metrizable space is Hausdorff right? Or you can say that the number of limit points should be less or equal to the number of converging sequences... -Kai
** Yes we use the fact that metric spaces are <math>T_2</math>. [[User:Drorbn|Drorbn]] 20:30, 26 October 2010 (EDT)

*Question. When proving topologist's sine curve is not path connected in page 157. Are we using the fact that for a cts function in Euclidean space if x_n converges to x then f(x_n) has to converge to f(x)? i.e. any open nbd of f(x) contains all but finitely many terms of f(x_n)? -Kai[[User:Xwbdsb|Xwbdsb]] 14:01, 27 October 2010 (EDT)
** We did not complete the proof of that fact in class. There are many ways to complete it, some use what you wrote and some don't. [[User:Drorbn|Drorbn]] 14:48, 27 October 2010 (EDT)
*Question. Is f(t) = (t,t,t...) cts in the R^\omega with uniform topology?.... -Kai [[User:Xwbdsb|Xwbdsb]] 15:23, 27 October 2010 (EDT)

10-327/Homework Assignment 4

2010-10-22T20:12:22Z

Ian: /* Questions */

10-327/Homework Assignment 4

2010-10-22T20:08:52Z

Ian: /* Questions */

10-327/Homework Assignment 4

2010-10-22T20:03:55Z

Ian:

10-327/Homework Assignment 4

2010-10-22T20:03:33Z

Ian: /* Questions */

10-327/Homework Assignment 4

2010-10-22T20:02:21Z

Ian: /* Questions */

10-327/Class Photo

2010-10-18T04:03:25Z

Ian: /* Who We Are... */

Our class on September 30, 2010:

[[Image:10-327-ClassPhoto.jpg|thumb|centre|800px|Class Photo: click to enlarge]]
{{10-327/Navigation}}

Please identify yourself in this photo! There are two ways to do that:

* [[Special:Userlogin|Log in]] to this Wiki and edit this page. Put your name, userid, email address and location in the picture in the alphabetical list below.
* Send [[User:Drorbn|Dror]] an email message with this information.

The first option is more fun but less private.

===Who We Are...===

{| align=center border=1 cellspacing=0
|-
!First name
!Last name
!UserID
!Email
!In the photo
!Comments
{{Photo Entry|last=Bar-Natan|first=Dror|userid=Drorbn|email=drorbn@ math.toronto.edu|location=facing everybody, as the photographer|comments=Take this entry as a model and leave it first. Otherwise alphabetize by last name. Feel free to leave some fields blank. For better line-breaking, leave a space next to the "@" in email addresses.}}



{{Photo Entry|last=Asher|first=Matt|userid=Asher|email=matt.asher utoronto.ca|location=[http://mattasher.com/images/Me_in_photo.jpg Under the arrow] |comments= }}

{{Photo Entry|last=Darwin|first=Ben|userid=bcd|email=[first.last] @ utoronto.ca|location=Back row, 2nd from left |comments= }}

{{Photo Entry|last=DiPoce|first=Christian|userid=Cdipoce|email=christian.dipoce@ utoronto.ca|location=very back row, 2nd from right |comments= black t-shirt, yellow writing; as handsome as ever.}}

{{Photo Entry|last=Dranovski|first=Anne|userid=Anne.d|email=a.dranovski@ utoronto.edu|location=front row, jean jacket. |comments= }}

{{Photo Entry|last=Hung|first=Ian|userid=Ian|email=ian.hung@ utoronto.ca|location=Right end of second row|comments= I'm wearing a white hoodie}}

{{Photo Entry|last=Kang|first=Soo Min|userid=soomin_kang|email=soomin.kang@ utoronto.ca|location=2nd row (from front), 2nd (from left)|comments= eyes closed, beige windbreaker}}

{{Photo Entry|last=Milcak|first=Juraj|userid=milcak|email=j.milcak @ utoronto.edu|location=frontmost, rightmost.|comments= }}

{{Photo Entry|last=Woodfine|first=Jason|userid=Jdw|email=jason(dot)woodfine (at) utoronto(dot)ca|location=3rd Row (from front), 4th (from left).|comments= Grey collared shirt, arms crossed, glasses.}}

{{Photo Entry|last=Yang|first=Kai|userid=xwbdsb|email=kai.b.yang@utoronto.ca|location=.|comments= Was planning to take the picture but was late because of washroom...}}

{{Photo Entry|last=Zhao|first=Frank|userid=Fzhao|email=frank.zhao@ utoronto.ca|location=2nd Row (from front), 6th (from left).|comments= The one in the blue shirt}}

|}