Abstract Nonsense

Crushing one theorem at a time

Just For Fun(Rudin’s Topology Section) Part II


13.

Problem: Form a compact subset of the real numbers (with the usual topology) which has countably many limit points.

Proof: Let \displaystyle \Omega=\left\{\frac{1}{2^n}\left(1-\frac{1}{m}\right):m,n\in\mathbb{N}\right\} which is tediously, but easily verified. Sorry folks, I’m not interested enough to Tex it all out. Comment me asking for a full solution if you so wish.

14.

Problem: Given an open cover (0,1) (with the usual topology) which has no finite subcover.

Proof: How about \displaystyle \left\{\left(0,1-\frac{1}{n}\right):n\in\mathbb{N}\right\}. So, we must first show that this in fact covers (0,1). So, let x\in(0,1) be arbitrary and let \delta=1-x. Then, we have that 0<1-x and so by the Archimedean principle there exists some n_0\in\mathbb{N} such that \displaystyle \frac{1}{n_0}<1-x \implies x<1-\frac{1}{n_0} and thus \displaystyle x\in\left(0,1-\frac{1}{n_0}\right). Now, to see that no finite subcover works, let \displaystyle \left\{\left(0,1-\frac{1}{n_1}\right),\cdots,\left(0,1-\frac{1}{n_m}\right)\right\} be any finite subclass of of \displaystyle \left\{\left(0,1-\frac{1}{n}\right)\right\}_{n\in\mathbb{N}}. Then, let N=\max\{n_1,\cdots,n_m\}. Now, it clearly follows that \displaystyle \left(0,\frac{1}{n_1}\right)\cup\cdots\cup\left(0,1-\frac{1}{n_m}\right)=\left(0,1-\frac{1}{N}\right) and so we mus merely show that there is some \displaystyle x\in (0,1)-\left(0,1-\frac{1}{N}\right). But, we merely need realize that \displaystyle 1-\frac{1}{N}<\frac{1}{2}\left(1+1-\frac{1}{N}\right)<1 and so the conclusion follows. \blacksquare

15.

Problem: Show that the intersection of arbitrarily many closed subsets  of a metric space with the finite intersection property  may be empty if they are just closed, or just bounded.

Proof: Just take \mathbb{R} for example. Then, the class \left\{\mathbb{R}-B_n(0)\right\}_{n\in\mathbb{N}} is a class of closed subsets of \mathbb{R} with the FIP but their intersection is empty. To see this we merely note that for each x\in\mathbb{R} we have that x\notin \mathbb{R}-B_{m}(x),\text{ }m=\left \lceil |x|\right\rceil +1

Also, if we just have bounded then the claim is also wildly false in \mathbb{R}. Take \left\{B_{\frac{1}{n}}(0)-\{0\}\right\}_{n\in\mathbb{N}}. This is a class of bounded subsets of \mathbb{R} with the FIP, but clearly \displaystyle \bigcap_{n\in\mathbb{N}}\left(B_{\frac{1}{n}}(0)-\{0\}\right)=\varnothing since choosing x\in\mathbb{R}-\{0\} we have that \displaystyle \frac{1}{n_0}<|x| for some n\in\mathbb{N} (by the Archimedean principle) and thus x\notin B_{\frac{1}{n_0}}(0)-\{0\}. \blacksquare

16.

Problem: Regard \mathbb{Q} as a metric space with d(p,q)=|p-q|. Let E=\left\{q\in\mathbb{Q}:2<q^2<3\right\}. Prove that a) E is closed in \mathbb{Q}, b) it is bounded in \mathbb{Q}, but it is not compact in \mathbb{Q}.

Proof:

a) We first prove a nice little lemma (which is almost immediate from theorem 2.30)

Lemma: Let \mathcal{M} be a metric space and \mathcal{N} a subspace (in the metric space sense, don’t get too ahead of yourself with topological terms…we haven’t yet proven they are equivalent), then E\subseteq\mathcal{N} is closed in it if and only if E=G\cap\mathcal{N} for some set G\subseteq\mathcal{M} which is closed in \mathcal{M}.

Proof: First suppose that E\subseteq\mathcal{N} is open, then \mathcal{N}-E is open and thus by theorem 2.30 we have that \mathcal{N}-E=\mathcal{N}\cap\left(\mathcal{M}-G\right) for some set G closed in \mathcal{M}. A little set theoretic self-convincing then shows that E=\mathcal{N}\cap G.

Conversely, suppose that E=\mathcal{N}\cap G for some set G closed in \mathcal{M}. Then, \mathcal{N}-E=\mathcal{N}\cap\left(\mathcal{M}-G\right) (once again with a little set theoretic magic…be sure to be mindful of what set theoretic difference means here!) and thus by theorem 2.30 again we see that this means that \mathcal{N}-E is open and thus E closed. \blacksquare

From this it immediately follows that E is closed in \mathbb{Q} since evidently E=\mathbb{Q}\cap[\sqrt{2},\sqrt{3}] and [\sqrt{2},\sqrt{3}] is closed in \mathbb{R}.

b) It is evidently bounded.

c) We prove that E is not compact in \mathbb{Q} by a roundabout but instructive way.

Lemma: Let X be a topological space and Y a subspace. Then, Z is a compact subspace of Y if and only if it’s a compact subspace of X.

Proof: First suppose that Z is a compact subspace of Y and let \left\{U_\alpha\cap Z\right\}_{\alpha\in\mathcal{A}} be an open cover Z as a subspace of X where U_\alpha is open in X. Then, since Y is a subspace of X it evidently follows that U_\alpha\cap Y is open in Y for every \alpha\in\mathcal{A}. Thus, it is evident that \left\{U_\alpha\cap Y\cap z\right\}_{\alpha\in\mathcal{A}} is an open cover of Z as a subspace of Y. So, by assumption it admits a finite subcover \left\{U_{\alpha_1}\cap Y\cap Z,\cdots,U_{\alpha_n}\cap Y\cap Z\right\} and so it readily follows that \left\{U_{\alpha_1}\cap Z,\cdots,U_{\alpha_n}\cap Z\right\} is a finite subcover of \left\{U_\alpha\cap Z\right\}_{\alpha\in\mathcal{A}}

Conversely, suppose that Z is compact as a subspace of X and \left\{U_{\alpha}\cap Z\right\}_{\alpha\in\mathcal{A}} is an open cover for Z as a subspace of Y. Then, since Y is a subspace of X we have that U_\alpha=V_\alpha\cap Y for some open V_\alpha in X. Thus, it is evident that \left\{V_\alpha\cap Z\right\}_{\alpha\in\mathcal{A}} is an open cover for Z as a subspace of X and so by assumption it admits a finite subcover \left\{V_{\alpha_1}\cap Z,\cdots,V_{\alpha_n}\cap Z\right\} from where a quick check shows then that \left\{U_{\alpha_1}\cap Z,\cdots,U_{\alpha_n}\cap Z\right\} is the desired finite subcover of \left\{U_\alpha\cap Z\right\}_{\alpha\in\mathcal{A}}. \blacksquare

From the above lemma we see that E will be compact as a subspace of \mathbb{Q} if and only if it is compact as a subspace of \mathbb{R}. But, it is not since evidently \sqrt{2} is a limit point of E but not a point of E, and thus E is not closed and thus not compact. \blacksquare

17.

Problem: Let E=\left\{x\in[0,1]:x=.a_1a_2a_3\cdots,\text{ }a_k=4,7\right\}\subseteq [0,1]; a) Is E countable? b) Is E dense in [0,1]? c) Is E dense in [0,1]? d) Is E perfect in [0,1]?

Proof:

a) No, E is not countable. It is evident that E\simeq\{0,1\}^{\mathbb{N}}\simeq\mathcal{P}(\mathbb{N})\simeq\mathbb{R}. The only one that may be slightly hazy is that \{0,1\}^{\mathbb{N}}\simeq\mathcal{P}(\mathbb{N}) but it is pretty routine to check that \eta:\{0,1\}^\mathbb{N}\to\mathcal{P}(\mathbb{N}):f\mapsto f^{-1}(\{1\}) is a bijection. (this is just the correspondence between subsets and indicator functions).

b) Of course it isn’t dense. It is clear that \left(0,\tfrac{1}{3}\right)\cap E=\varnothing.

c) It suffices from prior comment to show that it’s compact as a subspace, and by virtue of the Heine-Borel theorem we must merely show that E is closed and bounded. Boundedness is clear though since \displaystyle E\subseteq\left[\frac{4}{9},\frac{7}{9}\right]. So, suppose that x\in D(E) but x\notin E. Then, we have that x=.b_1b_2\cdots where b_k\ne4,7 for some k\in\mathbb{N}. But, note then that for any e\in E we have that |e-x|\geqslant\frac{1}{10^k} and thus B_{\frac{1}{10^k}}(x)\cap E=\varnothing which of course is a contradiction. Thus, the conclusion follows.

d) It suffices to prove (since we proved in c) that E is closed) that E\subseteq D(E). To do this let x\in E and let \delta>0 be given. Then, by the Archimedean principle there exists some n\in\mathbb{N} such that \displaystyle \frac{3}{10^n}<\delta and so if x=.a_1a_2\cdots then we have that

e=.b_1b_2\cdots,\text{ }b_k=\begin{cases} a_k \quad\text{if}\quad k\ne n \\ 4,7\quad\text{if}\quad k=n\end{cases}

Where obviously we choose 4,7 to be the opposite of what a_n is. Then, \displaystyle |e-x|=\frac{3}{10^n}<\delta and e\ne x. It follows that x\in D(E).

18.

Problem: Is there a non-empty perfect subset of \mathbb{R} (with the usual topology) which contains no rational values?

Proof: The answer is yes, but to say it right takes some measure theory. We know that \mu(\mathbb{Q})=0 (the usual measure) and so there exists countably many open intervals \mathcal{I}_n such that \displaystyle \mathbb{Q}\subseteq\bigcup_{n\in\mathbb{N}}\mathcal{I}_n and \displaystyle \sum_{n=1}^{\infty}\mu\left(\mathcal{I}_n\right)<1. So, we have that \displaystyle \mathbb{R}-\bigcup_{n\in\mathbb{N}}\mu\left(\mathcal{I}_n\right) is closed, non-empty (since it is non-zero measure). So, once again since it has positive measure it must be uncountable and by problem 28 it follows that the set of all condensation points of \displaystyle \mathbb{R}-\bigcup_{n\in\mathbb{N}}\mathcal{I}_n is perfect, and since each condensation point is a limit point it must be a point of \displaystyle \mathbb{R}-\bigcup_{n\in\mathbb{N}}\mathcal{I}_n and thus not a rational number. \blacksquare

19.

Problem: a) If A,B are disjoint closed sets in a metric space \mathcal{M} then they are separated. b) If A,B\subseteq\mathcal{M} are disjoint and open they are separated. c) Fix x_0\in\mathcal{M},\delta>0 and define A=\left\{y\in \mathcal{M}:d(x_0,y)<\delta\right\} and B=\left\{y\in\mathcal{M}:d(x_0,y)>\delta\right\}, and prove that A,B are separated. d) Prove that every connected metric space with more than one point  is uncountable.

Proof:

a) Clearly A,B are separated since \overline{A}\cap B=A\cap\overline{B}=A\cap B=\varnothing.

b) Suppose that A\cap\overline{B}\ne\varnothing then it must follow that there is some x\in A\cap D(B). But, since x\in D(B) every neighborhood N of x intersects B and thus intersects \mathcal{M}-A and thus x\notin A^{\circ}=A which is a contradiction. An identical argument works for the other cases.

c) Merely note that \varphi:\mathcal{M}\to \mathbb{R}:y\mapsto d(x_0,y) is continuous (since it’s the normal metric restricted to \{x_0\}\times \mathcal{M}\approx \mathcal{M}). And, that A=\varphi^{-1}((-\infty,\delta)),B=\varphi^{-1}((\delta,\infty)) which are disjoint and open. The conclusion follows from b)

d) Suppose that \mathcal{M} were countable and has more than one point, then so is \mathcal{M}\times \mathcal{M} and then so is \displaystyle d\left(\mathcal{M}\times \mathcal{M}\right)=\left\{d(x,y):x,y\in \mathcal{M}\right\}. So, choose x_0,y_0\in \mathcal{M} to be distinct and let \xi=d(x_0,y_0)>0. But, since (0,\xi)\subseteq\mathbb{R} is uncountable and d\left(\mathcal{M}\times\mathcal{M}\right) countable there exists some \delta\in (0,\xi)-d\left(\mathcal{M}\times\mathcal{M}\right). So, let A=\left\{y\in\mathcal{M}:d(x_0,y)<\delta\right\} and B=\left\{y\in\mathcal{M}:d(x_0,y)>\delta\right\}. Then, A,B are non-empty (0\in A and y_0\in B) disjoint and since d(x_0,y)\ne\delta,\text{ }y\in\mathcal{M} their union is equal to \mathcal{M}. It follows that \mathcal{M} is not connected. \blacksquare

20.

Problem: Are closures and interiors of connected sets always connected?

Proof: Closures are, we reprove this fact for general topological spaces.

Lemma: Let X be a topological space, E a connected subspace and E\subseteq G\subseteq\overline{E} then G is a connected subspace.

Proof: Evidently \text{cl}_G\text{ }E=G. So, let \varphi:G\to D (where D is the two-point discrete space) be continuous. Then since E is connected must have that \varphi\mid_{E} is constant, but since E is dense in G, \varphi continuous and D Hausdorff it follows that \varphi is constant. The conclusion follows. \blacksquare

The interior of connected subspaces is not always connected though. Consider the “dumbbell” shape B_{1}[-2]\cup[-1,1]\times\{0\}\cup B_{1}[2]. This is clearly connected (it is the union of three connected sets with non-empty intersection) but it’s interior is B_{1}(-2)\cup B_{1}(2) which is as classically disconnected as one can get. \blacksquare

Remark: It is interesting to note that the result is true for \mathbb{R}, since the only connected subsets are interiors which (as can easily be checked by case) the interiors of which are intervals.


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May 13, 2010 - Posted by | Analysis, Fun Problems, Topology, Uncategorized | , , , , ,

2 Comments »

  1. We could do 18 without measure theory: construct a Cantor set, and first enumerate all rationals as {q_n : n in N }.
    We then do the usual middle third construction, only at the n-th stage we make sure that q_n is not in the intermediate set at stage n (that we take the intersection of). We can do this quite easily. The result is a Cantor set that misses all of Q.

    Of course C == C x C (but this needs e.g. C == {0,1}^N) and so C is covered by uncountably many disjoint copies of the Cantor set, so plenty of them miss Q. This is a pure counting argument.

    Comment by Henno | May 13, 2010 | Reply


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