Math4201 Topology I (Lecture 28)

Consider $x\in X$, there is an element $U_\alpha$ in the open covering such that $x\in U_\alpha$.

In particular, there is $r_x$ such that $B_{r_x}(x)\subseteq U_\alpha$.

Then the collection $\{B_{\frac{r_x}{2}}(x)\}_{x\in X}$ is an open covering of $X$. (each $x\in X$ is contained in some $B_{\frac{r_x}{2}}(x)$)

Since $X$ is compact, there is a finite subcover $\{B_{\frac{r_{x_i}}{2}}(x_i)\}_{i=1}^n$ of $X$. Such that $\bigcup_{i=1}^n B_{\frac{r_{x_i}}{2}}(x_i)=X$.

Let $\delta = \min\{r_{\frac{r_{x_1}}{2}}, ..., r_{\frac{r_{x_n}}{2}}\}>0$.

Let $A\subseteq X$ be a subset with diameter less than $\delta$.

Take $y\in A$, then $A\subseteq B_\delta(y)$.

Take $x_i$ such that $y\in B_{\frac{r_{x_i}}{2}}(x_i)$. (such cover exists by definition of the subcover)

And then $\alpha$ such that $B_{\frac{r_{x_i}}{2}}(x_i)\subseteq U_\alpha$.

We claim that $B_\delta(y)\subseteq U_\alpha$, which would imply that $A\subseteq U_\alpha$.

$y\in B_{\frac{r_{x_i}}{2}}(x_i)$, and we know that $B_{r_{x_i}}(x_i)\subseteq U_\alpha$.

Since $\delta < \frac{r_{x_i}}{2}$, it suffices to show that $B_{\frac{r_{x_i}}{2}}(y)\subseteq U_\alpha$.

For any $z\in B_{\frac{r_{x_i}}{2}}(y)$, we have $d(z,y)<\frac{r_{x_i}}{2}$, and $d(y,x_i)<\frac{r_{x_i}}{2}$.

So $d(z,x_i)\leq d(z,y)+d(y,x_i)<\frac{r_{x_i}}{2}+\frac{r_{x_i}}{2}=r_{x_i}$, so $z\in B_{r_{x_i}}(x_i)\subseteq U_\alpha$.

So $B_{\frac{r_{x_i}}{2}}(y)\subseteq U_\alpha$.

Consider $X=(0,1)$ is not compact with the standard topology.

Consider $Z_n=(0,\frac{1}{n}]$, each interval is closed in $X$. This satisfies the finite intersection property because $\bigcap_{i=1}^k Z_{n_i}\neq \emptyset$ for any finite subcollection $\{Z_{n_1}, ..., Z_{n_k}\}$. We can find a smaller for any finite subcollection to get a non-empty intersection.

But $\bigcap_{n=1}^\infty Z_n = \emptyset$.

$\implies$

Let $U_\alpha=X-Z_\alpha$ is open for each $\alpha\in I$. By contradiction, suppose that $\bigcap_{\alpha\in I} Z_\alpha = \emptyset$.

So $X-\bigcap_{\alpha\in I} Z_\alpha = X=\bigcup_{\alpha\in I} U_\alpha = \bigcup_{\alpha\in I} (X-Z_\alpha)=X$.

So $\{U_\alpha\}_{\alpha\in I}$ is an open cover of $X$. Since $X$ is compact, there is a finite subcover $\{U_{\alpha_1}, ..., U_{\alpha_n}\}$.

So $\bigcap_{i=1}^n U_{\alpha_i} = X-\bigcup_{i=1}^n Z_{\alpha_i} = X$, $\bigcap_{i=1}^n Z_{\alpha_i} = \emptyset$. This contradicts the finite intersection property.

$\impliedby$

Proof is similar.

$X=[0,1]\cup \{2\}$ with subspace topology from $\mathbb{R}$.

Then $\{2\}$ is an isolated point $\{2\}=X\cap (2-\frac{1}{2}, 2+\frac{1}{2})$.

Proof by contradiction.

Let $X=\{x_n\}_{n\in\mathbb{N}}$ be a countable set.

Since $x_1$ is not an isolated point, so there exists $y_1\in X$ such that $y_1\neq x_1$. Apply the Hausdorff property, there exists disjoint open neighborhoods $U_1$ and $V_1$ such that $x_1\in U_1$ and $y_1\in V_1$.

In particular $\overline{V_1}$ does not contain $x_1$, but it contains $y_1$. (Follows from disjoint open neighborhoods)

Since $x_2$ is not an isolated point, so there exists $y_2\in X$ such that $y_2\neq x_2$. Apply the Hausdorff property, there exists disjoint open neighborhoods $U_2$ and $V_2$ such that $x_2\in U_2$ and $y_2\in V_2$.

If $x_2\notin V_1$, then we define $V_2$ as $V_1$.

If $x_2\in V_1$, then by the assumption, there is another point $y_2$ in $V_1$ which isn’t the same as $x_2$.

CONTINUE NEXT TIME.

It suffices to prove this for a closed interval $[a,b]$ with $a<b$. Because any interval contains such a closed interval.

The claim for a closed interval $[a,b]$ follows from the following theorem because $[a,b]$ is a non-empty compact Hausdorff space without an isolated point.