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. (In problem set 11)

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

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})