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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.