Difference between revisions of "10-327/Homework Assignment 4"

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*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
 
*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
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**Is it true that the closure of a path connected subspace is still path connected?..

Revision as of 19:21, 23 October 2010

Contents

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: dbnvp Topology-100927) in a manner similar to (say) dbnvp AKT-090910-1, and/or add links to the blackboard shots, in a manner similar to dbnvp 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 \le i \le 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?..