Math4201 Topology I (Lecture 24)

Connected and compact spaces

Connectedness

Recall from example last lecture, there exists a connected space but not path-connected space.

Lemma on connectedness

Let $X$ be a topological space and $A\subseteq X$ is a connected subspace. If $B\subseteq X$ satisfies $A\subseteq B\subseteq \overline{A}$ , then $B$ is connected. In particular, $\overline{A}$ is connected.

Proof

Assume that $B$ is not connected. In particular, there are open subspaces $U$ and $V$ of $X$ such that $U\cap B, V\cap B$ is a separation of $B$ .

Take $U\cap A, V\cap A$ , we show that this gives a separation of $A$ .

(i) Since $U,V$ are open, $U\cap A, V\cap A$ are open in $A$ .

(ii) Since $(U\cap B)\cap (V\cap B)=\emptyset$ , $(U\cap A)\cap (V\cap A)=\emptyset$ .

(iii) Since $(U\cap B)\cup (V\cap B)=B$ , any point in $B$ is in either $U\cap B$ or $V\cap B$ .

Since $A\subseteq B$ , $(U\cap A)\cup (V\cap A)=A$ .

(iv) $U\cap A$ and $V\cap A$ is nonempty by assumption $U\cap B$ is nonempty and contains $x\in B\cap U\subseteq \overline{A}$ . So any open neighborhood of $x$ have non-empty intersection $x'\in A$ , so $x'\in U\cap A$ and $U\cap A$ is nonempty. Similarly, $V\cap A$ is nonempty.

So $U\cap A$ and $V\cap A$ is a separation of $A$ , which contradicts the assumption that $A$ is connected.

Therefore, $B$ is connected.

Topologists’ sine curve

Let $A=\{(x,y)\in \mathbb{R}^2\mid y=\sin(1/x), x>0\}$ . Then $A$ is connected, and also path-connected.

\gamma(t) = (t, \sin(1/t)) \text{ for } t\in (0,1]

However, take $\overline{A}=A\cup \{0\}\times [-1,1]$ . Then $\overline{A}$ is not path-connected but connected.

Proof that topologists' sine curve is not path-connected

We want to show $X=\overline{A}$ has no continuous path

\gamma:([0,1])\to X

such that $\gamma(0)=(0,0)$ and $\gamma(1)=(1,\sin(1))$ .

If there exists such a path, let $t_0\in [0,1]$ be defined as

t_0=\sup\{t\in [0,1]\mid \gamma(t)=(0,x), x\in [0,1]\}

By the assumption on $t_0$ , we can find a sequence $\{t_n\}_{n\in\mathbb{N}_+}\subseteq A$ such that $t_n\to t$ .

By continuity of $\gamma$ , we have $\gamma(t_n)\to \gamma(t_0)$ , $(0,y_n)\to (0,y_0)$ .

Now focus on the restriction of $\gamma$ to $[t_0,1]$ , $\gamma:[t_0,1]\to X$ , $\gamma(t_0)=(0,y_0)$ , $\gamma(1)=(1,\sin(1))$ .

$t\in (t_0,1]$ , then $\gamma(t)\in$ graph of $y=\sin(1/x)$ .

Consider $\pi$ be the projection map to $x$ -axis, $\pi\circ \gamma:[t_0,1]\to \mathbb{R}$ , $\pi\circ \gamma(t_0)=0$ and $\pi\circ \gamma(1)=1$ .

In particular, there is a sequence $s_n\in [t_0,1]$ such that $s_n\to t_0$ and $\pi\circ \gamma(s_n)=\frac{1}{n\pi+\frac{\pi}{2}}$ . (using intermediate value theorem)

Then $\gamma(s_n)=(\frac{1}{n\pi},\sin(n\pi+\frac{\pi}{2}))=(\frac{1}{n\pi},(-1)^n)$ .

Since as $s_n\to t_0$ , and $\gamma$ is continuous, then we get a contradiction that the sequence $\gamma(s_n)$ should converge to $(0,t_0)$ where it is not.

Compactness

Motivation: in real numbers.

Extreme value theorem

Let $f:[a,b]\to \mathbb{R}$ be continuous. Then there are $x_m,x_M\in [a,b]$ such that $f(x_m)\leq f(x)\leq f(x_M)$ for all $x\in [a,b]$ .

Definition of cover

Let $X$ be a topological space. A covering of $X$ is a collection of subsets of $X$ that covers $X$ .

\{U_\alpha\}_{\alpha\in I}

such that $X=\bigcup_{\alpha\in I} U_\alpha$ .

An open cover of $X$ is a covering of $X$ such that each $U_\alpha$ is open.

Definition of compact space

A topological space $X$ is compact if for any open covering $\{U_\alpha\}_{\alpha\in I}$ of $X$ , there exists a finite subcovering $\{U_{\alpha_i}\}_{i=1}^n$ such that $X=\bigcup_{i=1}^n U_{\alpha_i}$ .

Example of non-compact space

Consider the interval $(0,1]$ , the open covering $(\frac{1}{n},1]$ open in $(0,1]$ , $\{(\frac{1}{n},1]\}_{n\in \mathbb{N}_+}$ has no finite subcovering.