Abstract Nonsense

Crushing one theorem at a time

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


Problem: Let A and B be separated subsets of \mathbb{R}^n, suppose that \bold{a}\in A,\bold{b}\in B and define

\bold{p}:\mathbb{R}\to\mathbb{R}^n:t\mapsto (1-t)\bold{a}+t\bold{b}

Let A_0=\bold{p}^{-1}(A),B_0=\bold{p}^{-1}(B). a) Prove that A_0,B_0 are separated subsets of \mathbb{R}. b) Prove that there exists t_0\in(0,1) such that \bold{p}(t_0)\notin A\cup B. c) Prove that every convex subset of \mathbb{R}^n is connected.


a) Clearly we have that A_0\cap B_0=\varnothing and so, if we assume that x\in\overline{A_0}\cap B_0 then we must have that x\in D(A_0)\cap B_0. So, we may choose \{x_n:n\in\mathbb{N}\}\subseteq A_0 such that x_n\to x and so it easily follows that \bold{p}(x_n)\to\bold{p}(x) and thus \bold{p}(x)\in\overline{A} so \bold{p}(x)\notin B and thud x\notin B_0 which is a contradiction.

b) We must merely note that if \bold{p}((0,1))\subseteq A\cup B then it would be a path from \bold{a}\in A to \bold{b}\in B which is impossible since they are separated.

c) We can prove this more generally. Let \mathcal{V} be a normed vector space and let C\subseteq\mathcal{V} be convex, then C is path connected. This clearly follows since the straight line L(\bold{x},\bold{y})=\left\{t\bold{x}+(1-t)\bold{y}:t\in[0,1]\right\} is a path from \bold{x} to \bold{y}.


Problem: A metric space \mathcal{M} is called separable if it contains a countable dense subset. Show that \mathbb{R}^n is separable.

Proof: We prove the much, much deeper following theorem.

Lemma: Let \left\{X_n\right\}_{n\in\mathbb{N}} be a countable class of non-empty separable topological spaces with corresponding countable dense subsets \left\{\mathfrak{D}_n\right\}_{n\in\mathbb{N}}. Then, \displaystyle X=\prod_{n\in\mathbb{N}}X_n is separable with the product topology.

Proof: For each n\in\mathbb{N} select an arbitrary but fixed \mathfrak{d}_n\in\mathfrak{D}_n and define \displaystyle \mathcal{D}_m=\prod_{n\in\mathbb{N}}D_n where

D_n=\begin{cases} \mathfrak{D}_n\quad\text{if}\quad n\leqslant m \\ \{\mathfrak{d}_n\}\quad\text{if}\quad n>m\end{cases}

Clearly we have that D_m\simeq\mathfrak{D}_1\times\cdots\times\mathfrak{D}_m and thus countable (the finite product of countable sets is countable). So, let

\displaystyle \mathfrak{D}=\bigcup_{m\in\mathbb{N}}\mathcal{D}_m

Then \mathfrak{D}, being the countable union of countable sets, is countable. We now claim that it is dense in X. To see this let \bold{x}\in X be arbitrary and \bold{N} any basic neighborhood of it. Then, there are only finitely many indices \{j_1,\cdots,j_k\} such that \pi_{j_k}(\bold{N})\ne X_{j_k}. So, let J=\max\{j_1,\cdots,j_k\}. Now, since for each k=1,\cdots,J we have that \pi_k(\bold{N}) is a neighborhood of \pi_k(\bold{x}) in X_k there is some d_k\in\mathfrak{D}_k\cap\pi_k(\bold{N}). So, doing this for k=1,\cdots,J we arrive at J points d_1,\cdots,d_J. So, then

\displaystyle \prod_{n\in\mathbb{N}}\{x_n\},\quad x_n=\begin{cases}d_n\quad\text{if}\quad n\leqslant J\\ \mathfrak{d}_n\quad\text{if}\quad n>J\end{cases}\in D_J\cap\bold{N}

Which proves that \mathfrak{D} is dense. The conclusion follows. \blacksquare

The above is kind of heavy duty (though it works since the product topology and usual topology coincide). For people who are more interested in analysis, the real way to do this is to note that \mathbb{Q}^n is countable and note that for any (x_1,\cdots,x_n)\in\mathbb{R}^n you can choose corresponding (q_1,\cdots,q_n)\in\mathbb{Q}^n such that |x_k-q_k|<\sqrt{\frac{\varepsilon^2}{4n}} and do the simple calculation.


Problem: A collection \left\{V_\alpha\right\}_{\alpha\in\mathcal{A}} of open subsets of \mathcal{M} ( a metric space) is said to be a base if for every x\in\mathcal{M} and every neighborhood U of it there is some V_\alpha such that x\in V_\alpha\subseteq U. Prove that every separable metric space has a countable base (in other words that it’s second countable).

Proof: Let \mathfrak{D} be the countable dense subset and let \mathfrak{B}=\left\{B_q(\xi):q\in\mathbb{Q},\text{ }\xi\in\mathfrak{D}\right\}. Clearly \mathfrak{B} is countable . To see this we merely note that \eta:\mathfrak{D}\times\mathbb{Q}\to\mathfrak{B}:(\xi,q)\mapsto B_q(\xi) is clearly a surjection. We shall now prove that \mathfrak{B} (as it’s letter suggests) is a base. So, let x\in\mathcal{M} be arbitrary and \varepsilon>0 be given. Since \mathfrak{D} is dense in \mathcal{M} there exists some \xi \in B_{\frac{\varepsilon}{2}}(x)\cap\mathfrak{D}. So, let \delta=d(x,\xi)<\frac{\varepsilon}{2} and choose q\in\mathbb{Q} such that \delta<q<\frac{\varepsilon}{2}. then, x\in B_q(\xi) and if y\in B_q(\xi) we have that d(y,x)\leqslant d(y,\xi)+d(\xi,x)<q+\frac{\varepsilon}{2}<\varepsilon and thus y\in B_q(\xi). The conclusion follows. \blacksquare


Problem: Let \mathcal{M} be a metric space in which every infinite subset of it has a limit point. Prove that \mathcal{M} is separable.

Proof: Consider the open cover \left\{B_\delta(x)\right\}_{x\in\mathcal{M}} and suppose that it did not have a finite subcover. Then, we may form a sequence \left\{x_n\right\}_{n\in\mathbb{N}} such that d(x_n,x_m)\geqslant \delta,\text{ }m<n, for otherwise if there existed some n_0\in\mathbb{N} such that for no x\in\mathcal{M} we have that d(x,x_{n_0}),\cdots,d(x,x_1)\geqslant \delta then for every x\in\mathcal{M} we have that

d(x,x_1)<\delta\text{ or }\cdots\text{ or }d(x,x_{n_0-1})<\delta

or in other words that x\in B_{\delta}(x_1)\cup\cdots\cup B_{\delta}(x_{n_0-1}) contradicting the assumption that \left\{B_{\delta}(x)\right\}_{x\in\mathcal{M}} has no finite subcover. But, let x\in\mathcal{M} and consider B_{\delta}(x) clearly B_{\delta}(x) contains at most one point of \left\{x_n:n\in\mathbb{N}\right\}. To see this suppose that it contained more than one point, say x_m,x_n\in B_{\delta}(x) then we may assume WLOG that m<n but this contradicts the construction of the sequence since d(x_m,x_n)\geqslant \delta. It follows that \left\{B_{\delta}(x)\right\}_{x\in\mathcal{M}} has a finite subcover.  So, let for each m\in\mathbb{N} we may cover \mathcal{M} with finitely many open balls of the form B_{\frac{1}{m}}(x). So, let A_m be the finite set of points such that B_{\frac{1}{m}}(x_1)\cup\cdots\cup B_{\frac{1}{m}}(x_n)=\mathcal{M} and let \displaystyle A=\bigcup_{m=1}^{\infty}A_m. Clearly A is countable and given any x\in\mathcal{N} and any neighborhood U of it we may (by the Archimedean principle) find some B_{\frac{1}{m}}(x) such that B_{\frac{1}{m}}(x)\subseteq U. But, since every point of \mathcal{M} is less than \frac{1}{m} in distance from some point of A_m it follows there is some \xi\in B_{\frac{1}{m}}\cap A_m. The conclusion follows. \blacksquare

Remark: All we really showed above (the hard part anyways) is that every limit point compact metric space is totally bounded. As we prove in the next theorem limit point compactness is implied by compactness in metric spaces (they are in fact equivalent) and thus every compact metric space is totally bounded. It actually is true that any space such that every \varepsilon>0 there is a countable \varepsilon-net is separable, this is done precisely as in the last part of the above.


Problem: Prove that every compact metric space \mathcal{M} has a countable base.

Proof: It suffices to show that every infinite subset of \mathcal{M} has a limit point, from where it will follow from 24. that it is separable, and thus second countable. So, let E\subseteq\mathcal{M} have no limit point. Then, for each x\in\mathcal{M} there exists some open ball B_{\delta_x}(x) such that E\cap B_{\delta_x}(x)\subseteq\{x\}. So, we may cover \mathcal{M} and so by assumption this open cover must admit a finite subcover \left\{B_{\delta_{x_1}}(x_1),\cdots,B_{\delta_{x_n}}(x_n)\right\}. So,

E=E\cap\mathcal{M}=\left(E\cap B_{\delta_{x_1}}(x_1)\right)\cup\cdots\cup\left(E\cap B_{\delta_{x_n}}(x_n)\right)\subseteq\left\{x_1,\cdots,x_n\right\}

The conclusion follows. \blacksquare


Problem: Let \mathcal{M} be a metric space such that every infinite subset has a limit point. Show that \mathcal{M} is compact.


Lemma: Let \mathcal{M} be a limit point compact metric space and \left\{F_n\right\}_{n\in\mathbb{N}} a sequence of descending (F_n\supseteq F_{n+1}) of non-empty closed subsets of \mathcal{M}. Then, \displaystyle \bigcap_{n\in\mathbb{N}}F_n\ne\varnothing

Proof: For each n\in\mathbb{N} choose some x_n\in F_n. We break this into two cases,  namely whether the set of all these x_n‘s is finite or infinite.

Case 1: Suppose that \left\{x_n:n\in\mathbb{N}\right\} is finite but \displaystyle \bigcap_{n\in\mathbb{N}}F_n=\varnothing. We first note that for each x\in F_1 there is some N_x\in\mathbb{N} such that N_x\leqslant n\implies x\notin U_n. To see this assume not and let n\in\mathbb{N} then x\in F_n otherwise x\notin F_n and thus x\notin F_{n+1},\cdots which was assumed to be impossible.  So, x\in F_n for all n\in\mathbb{N} and thus \displaystyle x\in\bigcap_{n\in\mathbb{N}}F_n which contradicts our assumption that it was empty. So, for each x\in \left\{x_n:\in\mathbb{N}\right\} there exists some N_x for which the above idea is true. Let N=\max\left\{N_x:x\in\left\{x_n:n\in\mathbb{N}\right\}\right\} then \left\{x_n:n\in\mathbb{N}\right\}\cap F_{N+1}=\varnothing which contradicts the construction of the set. It follows that the intersection \displaystyle \bigcap_{n\in\mathbb{N}}F_n is non-empty.

Case 2: Suppose that \left\{x_n:n\in\mathbb{N}\right\} is infinite. By assumption it then has a limit point \xi. So, we show that \displaystyle \xi\in\bigcap_{n\in\mathbb{N}}F_n by showing that it is contained in a fixed but arbitrary F_{n_0}. So, suppose that \xi\notin F_{n_0} then \xi\in \mathcal{M}-F_{n_0} which is open. But, since F_{n_0}\supseteq F_{n_0+1}\cdots it follows that \left(\mathcal{M}-F_{n_0}\right)\cap\left\{x_n:n\in\mathbb{N}\right\} is finite contradicting that \xi is a limit point of that set. It follows that \displaystyle \xi\in\bigcap_{n\in\mathbb{N}}F_n. \blacksquare

So, now let \Omega be an open cover for \mathcal{M}. Since \mathcal{M} is second countable we know that \Omega admits a countable subcover \left\{U_n\right\}_{n\in\mathbb{N}}. Now, suppose that U_1\cup\cdots\cup U_m\ne\mathcal{M} for every m\in\mathbb{N} then \left\{F_n\right\}_{n\in\mathbb{N}} where F_n=\mathcal{M}-\left(U_1\cup\cdots\cup U_n\right) is a sequence of decreasing, non-empty closed sets. But,

\displaystyle \bigcap_{n\in\mathbb{N}}F_n=\mathcal{M}-\bigcup_{n\in\mathbb{N}}\left(U_1\cup\cdots\cup U_n\right)=\mathcal{M}-\mathcal{M}=\varnothing

Which contradicts the lemma. The conclusion follows. \blacksquare


Problem: Let \mathcal{M} be a metric space and call \xi\in\mathcal{M} a condensation point of a set E\subseteq\mathcal{M} if every neighborhood of \xi contains uncountably many points of E. Now, suppose that E\subseteq\mathbb{R}^n is uncountable and let \mathfrak{C} be the set of all condensation points of E. a) Prove that \mathcal{C} is perfect and show that \left(\mathbb{R}-\mathfrak{C}\right)\cap E is countable.


a) Let x\in \mathfrak{C} but suppose that x\notin D(\mathfrak{C}) then there exists some neighborhood U-\{x\} of it such that for every y\in U there exists a neighborhood N_y of it such that \text{card }N_y\cap E\leqslant\aleph_0. So, we now need a quick lemma

Lemma: Let X be a second countable topological space and U any subset of X. Then, if \left\{V_\alpha\right\}_{\alpha\in\mathcal{A}} is a collection of open sets in X such that \displaystyle U\subseteq\bigcup_{\alpha\in\mathcal{A}}V_\alpha then there exists some countable subcollection \left\{V_{\alpha_n}\right\}_{n\in\mathbb{N}} such that \displaystyle U\subseteq\bigcup_{n\in\mathbb{N}}V_{\alpha_n}

Proof: For each u\in U there is some V_\alpha such that u\in V_\alpha but if \mathfrak{B} is the countable base there is some B\in\mathfrak{B} such that u\in B\subseteq V_{\alpha}. Doing this for each u\in U clearly gives a collection \left\{B_\beta\right\}_{\beta\in\mathcal{B}}\subseteq\mathfrak{B} such that \displaystyle U\subseteq\bigcup_{\beta\in\mathcal{B}}B_\beta. But, by design it is clear that \mathcal{B} is countable and so for each \beta\in\mathcal{B} we may choose some (since this is how they were begotten) some V_\beta\in\left\{V_\alpha\right\}_{\alpha\in\mathcal{A}} such that B_\beta\subseteq V_\beta. So, doing this for each \beta\in\mathcal{B} we arrive at a countable collection \left\{V_\beta\right\}_{\beta\in\mathcal{B}}\subseteq\left\{V_\alpha\right\}_{\alpha\in\mathcal{A}} such that

\displaystyle U\subseteq\bigcup_{\beta\in\mathcal{B}}B_\beta\subseteq\bigcup_{\beta\in\mathcal{B}}V_\beta

from where the conclusion follows. \blacksquare

So, note that since \mathbb{R}^n is second countable and we clearly have that \left\{N_y\cap Y\right\}_{y\in U} is a collection of open sets whose union contains U and so by the above lemma it follows that it must admit a countable subcover \left\{N_{y_n}\cap U\right\}_{n\in\mathbb{N}} so then

\displaystyle \text{card }U\cap E=\text{card }\bigcup_{n\in\mathbb{N}}\left(U\cap N_{y_n}\right)=\text{card }\bigcup_{n\in\mathbb{N}}\left(U\cap N_{y_n}\cap E\right)\leqslant \aleph_0

which contradicts that x\in \mathfrak{C}. So, this proves that \mathfrak{C}\subseteq D(\mathfrak{C}). Conversely, let x\in D(\mathfrak{C}) then every neighborhood U of x contains a point of \mathfrak{C} , but since U is also a neighborhood of that point it follows that \text{card }U\cap E>\aleph_0 from where it follows by the arbitrariness of U that x\in \mathfrak{C}.

b) Let \mathcal{W}=\left\{B\in\mathfrak{B}:\text{card }B\cap E\leqslant \aleph_0\right\}. We claim that \displaystyle \Sigma\overset{\text{def}}{=}\bigcup_{B\in\mathcal{W}}B=\mathbb{R}-\mathfrak{C}. To see this let x\in \Sigma then there is some B\in\mathfrak{B} such that \text{card }B\cap E\leqslant \aleph_0 and thus x\notin \mathfrak{C}. Conversely, let x\in\mathbb{R}-\mathfrak{C} then there exists some neighborhood U of x such that \text{card }U\cap E\leqslant\aleph_0 but there exists some B\in\mathfrak{B} such that x\in B\subseteq U. Thus, x\in\Sigma. So, we see that

\displaystyle \text{card }E\cap\Sigma=\text{card }E\cap\bigcup_{B\in\mathcal{W}}B=\text{card }\bigcup_{B\in\mathcal{W}}\left(B\cap E\right)\leqslant\aleph_0

The last part gotten from the fact that \mathcal{W}\subseteq\mathfrak{B} is countable and each E\cap B also countable. The conclusion follows. \blacksquare


May 14, 2010 - Posted by | Analysis, Fun Problems, Topology, Uncategorized | , , , ,

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