NoteNextra-origin/content/Math4201/Math4201_L35.md

Math4201 Topology I (Lecture 35)

Countability axioms

Kolmogorov classification

Consider the topological space X.

X is T_0 means for every pair of points x,y\in X, x\neq y, there is one of x and y is in an open set U containing x but not y.

X is T_1 means for every pair of points x,y\in X, x\neq y, each of them have a open set U and V such that x\in U and y\in V and x\notin V and y\notin U. (singleton sets are closed)

X is T_2 means for every pair of points x,y\in X, x\neq y, there exists disjoint open sets U and V such that x\in U and y\in V. (Hausdorff)

X is T_3 means that X is regular: for any x\in X and any close set A\subseteq X such that x\notin A, there are disjoint open sets U,V such that x\in U and A\subseteq V.

X is T_4 means that X is normal: for any disjoint closed sets, A,B\subseteq X, there are disjoint open sets U,V such that A\subseteq U and B\subseteq V.

Example

Let \mathbb{R}_{\ell} with lower limit topology.

\mathbb{R}_{\ell} is normal since for any disjoint closed sets, A,B\subseteq \mathbb{R}_{\ell}, x\in A and B is closed and doesn't contain x. Then there exists \epsilon_x>0 such that [x,x+\epsilon_x)\subseteq A and does not intersect B.

Therefore, there exists \delta_y>0 such that [y,y+\delta_y)\subseteq B and does not intersect A.

Let U=\bigcup_{x\in A}[x,x+\epsilon_x) is open and contains A.

V=\bigcup_{y\in B}[y,y+\delta_y) is open and contains B.

We show that U and V are disjoint.

If U\cap V\neq \emptyset, then there exists x\in A and Y\in B such that [x,x+\epsilon_x)\cap [y,y+\delta_y)\neq \emptyset.

This is a contradiction since [x,x+\epsilon_x)\subseteq A and [y,y+\delta_y)\subseteq B.

Theorem Every metric space is normal

Use the similar proof above.

Proof

Let A,B\subseteq X be closed.

Since B is closed, for any x\in A, there exists \epsilon_x>0 such that B_{\epsilon_x}(x)\subseteq B.

Since A is closed, for any y\in B, there exists \delta_y>0 such that A_{\delta_y}(y)\subseteq A.

Let U=\bigcup_{x\in A}B_{\epsilon_x/2}(x) and V=\bigcup_{y\in B}B_{\delta_y/2}(y).

We show that U and V are disjoint.

If U\cap V\neq \emptyset, then there exists x\in A and Y\in B such that B_{\epsilon_x/2}(x)\cap B_{\delta_y/2}(y)\neq \emptyset.

Consider z\in B_{\epsilon_x/2}(x)\cap B_{\delta_y/2}(y). Then d(x,z)<\epsilon_x/2 and d(y,z)<\delta_y/2. Therefore d(x,y)\leq d(x,z)+d(z,y)<\epsilon_x/2+\delta_y/2.

If \delta_y<\epsilon_x, then d(x,y)<\delta_y/2+\delta_y/2=\delta_y. Therefore x\in B_{\delta_y}(y)\subseteq A. This is a contradiction since U\cap B=\emptyset.

If \epsilon_x<\delta_y, then d(x,y)<\epsilon_x/2+\epsilon_x/2=\epsilon_x. Therefore y\in B_{\epsilon_x}(x)\subseteq B. This is a contradiction since V\cap A=\emptyset.

Therefore, U and V are disjoint.

Lemma fo regular topological space

X is regular topological space if and only if for any x\in X and any open neighborhood U of x, there is open neighborhood V of x such that \overline{V}\subseteq U.

Lemma of normal topological space

X is a normal topological space if and only if for any A\subseteq X closed and any open neighborhood U of A, there is open neighborhood V of A such that \overline{V}\subseteq U.

Proof

\implies

Let A and U are given as in the statement.

So A and (X-U) are disjoint closed.

Since X is normal and A\subseteq V\subseteq X and V\cap W=\emptyset. X-U\subseteq W\subseteq X. where W is open in X.

And \overline{V}\subseteq (X-W)\subseteq U.

And A\subseteq V.

The proof of reverse direction is similar.

Let A,B be disjoint and closed.

Then A\subseteq U\coloneqq X-B\subseteq X and X-B is open in X.

Apply the assumption to find A\subseteq V\subseteq X and V is open in X and \overline{V}\subseteq U\coloneqq X-B.

Proposition of regular and Hausdorff on subspaces

1. If X is a regular topological space, and Y is a subspace. Then Y with induced topology is regular. (same holds for Hausdorff)
2. If \{X_\alpha\} is a collection of regular topological spaces, then their product with the product topology is regular. (same holds for Hausdorff)

Caution

The above does not hold for normal.

Recall that \mathbb{R}_{\ell} with lower limit topology is normal. But \mathbb{R}_{\ell}\times \mathbb{R}_{\ell} with product topology is not normal.

Proof that Sorgenfrey plane is not normal

The goal of this problem is to show that \mathbb{R}^2_\ell (the Sorgenfrey plane) is not normal. Recall that \mathbb{R}_\ell is the real line with the lower limit topology, and \mathbb{R}_\ell^2 is equipped with the product topology. Consider the subset

L = \{\, (x,-x) \mid x\in \mathbb{R}_\ell \,\} \subset \mathbb{R}^2_\ell.

Let A\subset L be the set points of the form (x,-x) such that x is rational and B\subset L be the set points of the form (x,-x) such that x is irrational.

1. Show that the subspace topology on L is the discrete topology. Conclude that A and B are closed subspaces of \mathbb{R}_\ell^2

Proof

First we show that $L$ is closed.

Consider x=(a,b)\in\mathbb{R}^2_\ell-L, by definition a\neq -b.

If a>-b, then there exists open neighborhood U_x=[\frac{\min\{a,b\}}{2},a+1)\times[\frac{\min\{a,b\}}{2},b+1) that is disjoint from L (no points of form (x,-x) in our rectangle), therefore x\in U_x.

If a<-b, then there exists open neighborhood U_x=[a,a+\frac{\min\{a,b\}}{2})\times[b,b+\frac{\min\{a,b\}}{2}) that is disjoint from L, therefore x\in U_x.

Therefore, \mathbb{R}^2_\ell-L=\bigcup_{x\in \mathbb{R}_\ell^2-L} U_x is open in \mathbb{R}^2_\ell.

So L is closed in \mathbb{R}^2_\ell.

To show L with subspace topology on \mathbb{R}^2_\ell is discrete topology, we need to show that every singleton of L is open in L.

For each \{(x,-x)\}\in L, [x,x+1)\times [-x,-x+1) is open in \mathbb{R}_\ell^2 and \{(x,-x)\}=([x,x+1)\times [-x,-x+1))\cap L, therefore \{(x,-x)\} is open in L.

Since A,B are disjoint and A\cup B=L, therefore A=L-B and B=L-A, by definition of discrete topology, A,B are both open therefore the complement of A,B are closed. So A,B are closed in L.

since L is closed in \mathbb{R}^2_\ell, by \textbf{Lemma \ref{closed_set_close_subspace_close}}, A,B is also closed in \mathbb{R}_\ell^2. Therefore A,B are closed subspace of \mathbb{R}_\ell^2.

2. Let V be an open set of \mathbb{R}^2_\ell containing B. Let K_n consist of all irrational numbers x\in [0,1] such that [x, x+1/n) \times [-x, -x+1/n) is contained in V. Show that [0,1] is the union of the sets K_n and countably many one-point sets.

Proof

Since $B$ is open in $L$, for each $b\in B$, by definition of basis in $\mathbb{R}_\ell^2$, and $B$ is open, there exists $b\in ([b,b+\epsilon)\times [-b,-b+\delta))\cap L\subseteq V$ and $0<\epsilon,\delta$, so there exists $n_b$ such that $\frac{1}{n_b}<\min\{\epsilon,\delta\}$ such that $b\in ([b,b+\frac{1}{n_b})\times [-b,-b+\frac{1}{n_b}))\cap L\subseteq V$.

Therefore \bigcup_{n=1}^\infty K_n covers irrational points in [0,1]

Note that B=L-A where A is rational points therefore countable.

So [0,1] is the union of the sets K_n and countably many one-point sets.

3. Use Problem 5-3 to show that some set \overline{K_n} contains an open interval (a,b) of \mathbb{R}. (You don't need to prove Problem 5.3, if it is not your choice of #5.)

Lemma

Let X be a compact Hausdorff space; let \{A_n\} be a countable collection of closed sets of X. If each sets A_n has empty interior in X, then the union \bigcup_{n=1}^\infty A_n has empty interior in X.

Proof

We proceed by contradiction, note that $[0,1]$ is a compact Hausdorff space since it's closed and bounded.

And \{\overline{K_n}\}_{n=0}^\infty is a countable collection of closed sets of [0,1].

Suppose for the sake of contradiction, \overline{K_n} has empty interior in X for all n\in \mathbb{N}, by \textbf{Lemma \ref{countable_closed_sets_empty_interior}}, then \bigcup_{n=1}^\infty \overline{K_n} has empty interior in [0,1], where \Q\cap[0,1] are countably union of singletons, therefore has empty interior in [0,1].

Therefore \bigcup_{n=1}^\infty \overline{K_n} has empty interior in [0,1], since \bigcup_{n=1}^\infty K_n\subseteq \bigcup_{n=1}^\infty \overline{K_n}, \bigcup_{n=1}^\infty K_n also has empty interior in [0,1] by definition of subspace of [0,1], therefore \bigcup_{n=1}^\infty K_n\cup (\Q\cap[0,1]) has empty interior in [0,1]. This contradicts that \bigcup_{n=1}^\infty K_n\cup (\Q\cap[0,1]) covers [0,1] and should at least have interior (0.1,0.9).

4. Show that V contains the open parallelogram consisting of all points of the form

x\times (-x+\epsilon)\quad\text{ for which }\quad a<x<b\text{ and }0<\epsilon<\frac{1}{n}.

Proof

Since $V$ is open, by previous problem we know that there exists $n$ such that $\overline{K_n}$ contains the open interval $(a,b)$.

If x\in K_n, \forall a<x<b, by definition of K_n [x,x+\frac{1}{n})\times [-x,-x+\frac{1}{n})\subseteq V.

If x is a limit point of K_n, since V is open, there exists 0<\epsilon<\frac{1}{n} such that [x,x+\epsilon)\times [-x,-x+\epsilon)\subseteq V.

This gives our desired open parallelogram.

5. Show that if q is a rational number with a<q<b, then the point q\times (-q) of \mathbb{R}_\ell^2 is a limit point of V. Conclude that there are no disjoint open neighborhoods U of A and V of B.

Proof

Consider all the open neighborhood of $q\times (-q)$ in $\mathbb{R}_\ell^2$, for all $\delta>0$, $[q,q+\delta)\times (-q,-q+\delta)$ will intersect with some $x\times [-x,-x+\epsilon)\subseteq V$ such that $0<\epsilon<\frac{1}{n}<\delta$.

Therefore, any open set containing q\times (-q)\in A will intersect with V, it is impossible to build disjoint open neighborhoods U of A and V of B.

This shows that \mathbb{R}_{\ell} is not metrizable. Otherwise \mathbb{R}_{\ell}\times \mathbb{R}_{\ell} would be metrizable. Which could implies that \mathbb{R}_{\ell} is normal.

Theorem of metrizability (Urysohn metirzation theorem)

If X is normal and second countable, then X is metrizable.

Note

• Every metrizable topological space is normal.
• Every metrizable space is first countable.
• But there are some metrizable space that is not second countable.

Note that if X is normal and first countable, then it is not necessarily metrizable. (Example \mathbb{R}_{\ell})