10-327/Homework Assignment 4: Difference between revisions
| Line 38: | Line 38: | ||
**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. |
**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}. |
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> |
|||
*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) |
|||
Revision as of 08:58, 23 October 2010
| ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
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 Video: 
Topology-100927) in a manner similar to (say) AKT-090910-1, and/or add links to the blackboard shots, in a manner similar to 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".
| Dror's notes above / Student's notes below |
Questions
1)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.
2) 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]\displaystyle{ F :\tilde R^n \rightarrow R^n }[/math] be defined as [math]\displaystyle{ F(x)= \prod_{i=1}^{n} \pi_i (x) }[/math] and let [math]\displaystyle{ F^{-1} : R^n \rightarrow \tilde R^n }[/math] be defined as [math]\displaystyle{ F^{-1}(x)= \prod_{i \in Z_+} f_i (x) }[/math] where [math]\displaystyle{ f_i (x) = \pi_i (x) }[/math] if [math]\displaystyle{ 1 \le i \le n }[/math] and [math]\displaystyle{ f_i(x)=0 }[/math] otherwise. Then both [math]\displaystyle{ F }[/math] and [math]\displaystyle{ 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]\displaystyle{ F }[/math] is a bijection because [math]\displaystyle{ F(F^{-1}(x_1, \ldots, x_n))=(x_1, \ldots, x_n) }[/math] and [math]\displaystyle{ F^{-1}(F(x_1, \ldots, x_n, 0,0, \ldots))=(x_1, \ldots, x_n, 0,0, \ldots) }[/math], i.e [math]\displaystyle{ F }[/math] has a left and right inverse. So [math]\displaystyle{ F }[/math] is a homeomorphism between the two spaces. Quick question is there a nicer way of writing math than using the math tag? Ian 16:03, 22 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 Xwbdsb 19:19, 22 October 2010 (EDT)
- If I understand your question, I don't see why not...think about [math]\displaystyle{ \mathbb{R} }[/math] for example. [math]\displaystyle{ f(x)=x^2 }[/math] is not onto, then let [math]\displaystyle{ g(x)=e^x }[/math] then g compose f is [math]\displaystyle{ 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
- 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.
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]\displaystyle{ S }[/math] is closed in [math]\displaystyle{ [0,1]. \Rightarrow S^C }[/math] is open. If [math]\displaystyle{ sup(S)=\alpha \notin S \Rightarrow \exists r\gt 0 }[/math] s.t. [math]\displaystyle{ B(\alpha, r) \subset S^C \Rightarrow \alpha - 0.5r \in S^C \Rightarrow \alpha - 0.5r \lt \alpha }[/math] is an upper bound for S. [math]\displaystyle{ \Rightarrow \Leftarrow }[/math]
- 2. Recall that [math]\displaystyle{ G = \{g | [0,g] \subset A\}; g_0 = sup(G) \Rightarrow \forall g \lt g_0, [0,g] \subset A \Rightarrow [0, g_0) \subset A }[/math]. So, if [math]\displaystyle{ g_0 \in A \Rightarrow [0,g_0] \subset A }[/math]. -Frank 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]\displaystyle{ \alpha - 0.5r }[/math] not in S, but neither is anything in [math]\displaystyle{ B(\alpha,r) }[/math], since [math]\displaystyle{ S^C }[/math] is open. There can be no elements [math]\displaystyle{ \geq \alpha }[/math] in S because it's the supremum. Recall also we're working in the Reals. [math]\displaystyle{ \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]\displaystyle{ \forall r \gt 0, \exists g \in G \cap B(g_0,r) \Rightarrow [0,g] \subset A \Rightarrow [0,g'] \subset A, \forall g' \lt g \Rightarrow g' \in G \Rightarrow \forall g \lt g_0 }[/math], take [math]\displaystyle{ 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]
- 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 Fzhao 09:58, 23 October 2010 (EDT)
