10-327/Homework Assignment 4

DBN: Classes: 2010-11: 10-327/Navigation Additions to the MAT 327 web site no longer count towards good deed points # Week of... Notes and Links 1 Sep 13 About This Class, Monday - Continuity and open sets, Thursday - topologies, continuity, bases. 2 Sep 20 Monday - More on bases, Thursdsay - Products, Subspaces, Closed sets, HW1, HW1 Solutions 3 Sep 27 Monday - the Cantor set, closures, Thursday, Class Photo, HW2, HW2 Solutions 4 Oct 4 Monday - the axiom of choice and infinite product spaces, Thursday - the box and the product topologies, metric spaces, HW3, HW3 Solutions 5 Oct 11 Monday is Thanksgiving. Thursday - metric spaces, sequencial closures, various products. Final exam's date announced on Friday. 6 Oct 18 Monday - connectedness in ${\mathbb {R} }$, HW4, HW4 Solutions, Thursday - connectedness, path-connectedness and products 7 Oct 25 Monday - Compactness of $[0,1]$, Term Test on Thursday, TT Solutions 8 Nov 1 Monday - compact is closed and bounded, maximal values, HW5, HW5 Solutions, Wednesday was the last date to drop this course, Thursday - compactness of products and in metric spaces, the FIP 9 Nov 8 Monday-Tuesday is Fall Break, Thursday - Tychonoff and a taste of Stone-Cech, HW6, HW6 Solutions 10 Nov 15 Monday - generalized limits, Thursday - Normal spaces and Urysohn's lemma, HW7, HW7 Solutions 11 Nov 22 Monday - $T_{3.5}$ and $I^{A}$, Thursday - Tietze's theorem 12 Nov 29 Monday - compactness in metric spaces, HW8, HW8 Solutions, Thursday - completeness and compactness 13 Dec 6 Monday - Baire spaces and no-where differentiable functions, Wednesday - Hilbert's 13th problem; also see December 2010 Schedule R Dec 13 See December 2010 Schedule F Dec 20 Final exam, Monday December 20, 2PM-5PM, at BR200 Register of Good Deeds Add your name / see who's in! See Hilbert's 13th

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.
   • The term test will cover everything including Monday October 25 and including this assignment. Drorbn 13:42, 23 October 2010 (EDT)
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 $F:{\tilde {R}}^{n}\rightarrow R^{n}$ be defined as $F(x)=\prod _{i=1}^{n}\pi _{i}(x)$ and let $F^{-1}:R^{n}\rightarrow {\tilde {R}}^{n}$ be defined as $F^{-1}(x)=\prod _{i\in Z_{+}}f_{i}(x)$ where $f_{i}(x)=\pi _{i}(x)$ if $1\leq i\leq n$ and $f_{i}(x)=0$ otherwise. Then both $F$ and $F^{-1}$ are continuous, because we are working in the product topology and the component functions, namely the projection function and constant function are continuous. Also $F$ is a bijection because $F(F^{-1}(x_{1},\ldots ,x_{n}))=(x_{1},\ldots ,x_{n})$ and $F^{-1}(F(x_{1},\ldots ,x_{n},0,0,\ldots ))=(x_{1},\ldots ,x_{n},0,0,\ldots )$, i.e $F$ has a left and right inverse. So $F$ 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 $\mathbb {R}$ for example. $f(x)=x^{2}$ is not onto, then let $g(x)=e^{x}$ then g compose f is $e^{x^{2}}$ - 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 $S$ is closed in $[0,1].\Rightarrow S^{C}$ is open. If $sup(S)=\alpha \notin S\Rightarrow \exists r>0$ s.t. $B(\alpha ,r)\subset S^{C}\Rightarrow \alpha -0.5r\in S^{C}\Rightarrow \alpha -0.5r<\alpha$ is an upper bound for S. $\Rightarrow \Leftarrow$
• 2. Recall that $G=\{g|[0,g]\subset A\};g_{0}=sup(G)\Rightarrow \forall g<g_{0},[0,g]\subset A\Rightarrow [0,g_{0})\subset A$. So, if $g_{0}\in A\Rightarrow [0,g_{0}]\subset A$. -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 $\alpha -0.5r$ not in S, but neither is anything in $B(\alpha ,r)$, since $S^{C}$ is open. There can be no elements $\geq \alpha$ in S because it's the supremum. Recall also we're working in the Reals. $\Rightarrow [\alpha -0.5r,\infty )\subset S^{C}.$

• For the second question, notice that the supremum is the least upper bound (of G), so $\forall r>0,\exists g\in G\cap B(g_{0},r)\Rightarrow [0,g]\subset A\Rightarrow [0,g']\subset A,\forall g'<g\Rightarrow g'\in G\Rightarrow \forall g<g_{0}$, take $r=0.5(g_{0}-g)\Rightarrow \exists a\in (g_{0}-r,g_{0})\cap G,\Rightarrow [0,g]\subset [0,a]\subset A.$

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 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
  • $S^{1}\subset {\mathbb {R} }^{2}$, and ${\mathbb {R} }^{2}$ is a vector space. 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 $B$ doesn't intersect $A$ it's complement will. Drorbn 13:42, 23 October 2010 (EDT)

• Hi Dror. Can you possibly give a little bit more hints on 9? I really still have no clue how to solve it although I have been thinking about it for the whole day. -Kai

• • Is it true that the closure of a path connected subspace is still path connected?.. It is true in Metric spaces with metric topology but how about general topological spaces? -Kai Xwbdsb 21:16, 23 October 2010 (EDT)