NoteNextra-origin/content/Math4201/Math4201_L36.md

Math4201 Topology I (Lecture 36)

Countable Axioms and Separation Axioms

Separation Axioms

Proposition for T_1 spaces

If X is a topological space such that all singleton are closed, then then following holds:

• X is regular if and only if for any point x\in X and any open neighborhood V of X such that \overline{V}\subseteq U.
• X is normal if and only if for any closed set A\subseteq X, there is an open neighborhood V of A such that \overline{V}\subseteq U.

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)

Proof

1. If X is regular and Y\subseteq X,then we want to show that Y is regular.

We take a point y\in Y and a closed subset A of Y which doesn't contain y.

Observe that A\subseteq Y is closed if and only if A=\overline{A}\cap Y where \overline{A} is the closure of A in X.

Notice that y\in Y but y\notin A, then y\notin \overline{A} (otherwise y\in A).

By regularity of X, we can find an open neighborhood y\in U\subseteq X and \overline{A}\subseteq V\subseteq X and U,V are open and disjoint.

This implies that U\cap Y and V\cap Y are open neighborhood of y and A and disjoint from each other.

(Proof for Hausdorff is similar)

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2. If \{X_\alpha\} is a collection of regular topological spaces, then their product with the product topology is regular.

We take a collection of regular spaces \{X_\alpha\}_{\alpha\in I}.

We want to show that their product with the product topology is regular.

Take \underline{x}=(x_\alpha)_{\alpha\in I}\in \prod_{\alpha\in I}X_\alpha, any open neighborhood of x contains a basis element of the form

\prod_{\alpha\in I}U_\alpha

with x_\alpha\in U_\alpha for all \alpha\in I, and all but finitely many U_\alpha are equal to X_\alpha. (By definition of product topology)

Now for each \alpha_i take an open neighborhood V_\alpha of x_{\alpha_i} such that

1. \overline{V_\alpha}\subseteq U_\alpha (This can be cond by regularity of X_\alpha)
2. V_\alpha=X_\alpha if U_\alpha=X_\alpha

The product of \prod_{\alpha\in I}V_\alpha is an open neighborhood of \underline{x} and

\overline {\prod_{\alpha\in I}V_\alpha}=\prod_{\alpha\in I}\overline{V_\alpha}\subseteq \prod_{\alpha\in I}U_\alpha

Therefore, X is regular.

(Proof for Hausdorff is similar)

If we replace the regularity with Hausdorffness, then we can take two points \underline{x},\underline{y}. Then there exists \alpha_0 such that x_{\alpha_0}\neq y_{\alpha_0}. We can use this to build disjoint open neighborhoods

x_{\alpha_0}\in U_{\alpha_0}\subseteq X_{\alpha_0},\quad y_{\alpha_0}\in V_{\alpha_0}\subseteq X_{\alpha_0}

of x_{\alpha_0} and y_{\alpha_0}.

Then we take U_\alpha=\begin{cases} U_{\alpha_0} &\alpha= \alpha_0\\ X_{\alpha_0} & \text{otherwise}\end{cases} and V_\alpha=\begin{cases} V_{\alpha_0} &\alpha= \alpha_0\\ X_{\alpha_0} & \text{otherwise}\end{cases}.

These two sets are disjoint and \prod_{\alpha\in I}U_{\alpha} and \prod_{\alpha\in I}V_{\alpha} are open neighborhoods of x and y.

Recall that this property does not hold for subspace of normal spaces.

How we construct new normal spaces from existing one

Theorem for constructing normal spaces

1. Any compact Hausdorff space is normal
2. Any regular second countable space is normal

Proof

For the first proposition

Earlier we showed that any compact Hausdorff space X is regular, i.e., for any closed subspace A of X and a point x\in X not in A. There are open neighborhoods U_x of A and V_x of x such that U_x\cap V_x=\emptyset.

Now let B be a closed subset of X disjoint from A.

For any x\in B, we know that we have a disjoint open neighborhood U_x of A and V_x of x.

\{V_n\}_{n\in \mathbb{B}} gives an open covering of B, B is closed subset of a compact space, therefore, B is compact.

This implies that \exists x_1,x_2,\ldots,x_n such that B\subseteq \bigcup_{i=1}^n V_{x_i}.

U\coloneqq \bigcup_{i=1}^n U_{x_i} is an open covering of A.

U_{x_i}\cap V_{x_i}=\emptyset implies that \bigcup_{i=1}^n U_{x_i}\cap \bigcup_{i=1}^n V_{x_i}=\emptyset.

In summary, U and V are disjoint open neighborhoods of A and B respectively.

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We want to show that any regular second countable space is normal.

Take A,B be two disjoint closed subsets of X. We want to show that we can find disjoint open neighborhoods U of A and V of B such that U\cap V=\emptyset.

Step 1:

There is a countable open covering \{U_i\}_{i\in I} of A such that for any i, \overline{U_i}\cap B=\emptyset.

For any x\in A, X-B is an open neighborhood of x and by reformulation of regularity, we can find an open set U_i' such that \overline{U_i'}\subseteq X-B.

(We will use the second countability of X to produce countable open coverings)

Let B be a countable basis and let U_x\in \mathcal{B} such that x\in U_x\subseteq U_x'.

Note that \overline{U_x}\subseteq \overline{U_x'}\subseteq X-B.

and \{U_x\}_{x\in A}\subseteq \mathcal{B} is a countable collection.

So this is a countable open covering of A by relabeling the elements of this open covering we can denote it by \{U_i\}_{i\in\mathbb{N}}.

Step 2:

There is a countable open covering \{V_i\}_{i\in I} of B such that for any i, \overline{V_i}\cap A=\emptyset.

Step 3:

Let's define \hat{U}_i=U_i-\bigcup_{j=1}^i\overline{V}_j=U_i\cap (X-\bigcup_{j=1}^i \overline{V}_j), note that (X-\bigcup_{j=1}^i \overline{V}_j) and U_i is open in X, therefore \hat{U}_i is open in X.

Since A\subseteq \bigcup_{i=1}^\infty U_i and \overline{V_j}\cap A=\emptyset, we have \bigcup_{i=1}^\infty \hat{U}_i\supseteq A.

Similarly, we have:

\hat{V}_i=V_i-\bigcup_{j=1}^i\overline{U}_j=V_i\cap (X-\bigcup_{j=1}^i \overline{U}_j)

is also open and \bigcup_{i=1}^\infty \hat{V}_i\supseteq B.

Then we claim that these two open neighborhoods U=\bigcup_{i=1}^\infty \hat{U}_i and V=\bigcup_{i=1}^\infty \hat{V}_i are disjoint.

To see this, we proceed by contradiction, suppose x\in \bigcup_{i=1}^\infty \hat{U}_i\cap \bigcup_{i=1}^\infty \hat{V}_i, x\in \bigcup_{i=1}^\infty \hat{U}_i and x\in \bigcup_{i=1}^\infty \hat{V}_i.

x\in \bigcup_{i=1}^\infty \hat{U}_i implies that \exists k such that x\in \hat{U}_k and \exists l such that x\in \hat{V}_l.

Suppose without loss of generality that k\leq l.

Then x\in \hat{U}_k\subseteq U_k, and x\in \hat{V} implies that \forall j\leq l:x\notin U_j. This gives x\notin U_k.

This is a contradiction.

Therefore, U and V are disjoint.