Compactness & Connectedness
Compactness & Connectedness
Compactness is the key topological property that generalizes 'closed and bounded' from $\mathbb{R}^n$. A space is compact if every open cover has a finite subcover. Continuous images of compact sets are compact; continuous bijections from compact to Hausdorff spaces are homeomorphisms. Connectedness captures the intuition that a space 'cannot be split into two disjoint open pieces.' The intermediate value theorem and extreme value theorem are purely topological consequences.
Compactness
$(X,\mathcal{T})$ is compact if every open cover $X=\bigcup_{\alpha}U_\alpha$ has a finite subcover $X=U_1\cup\cdots\cup U_n$. In metric spaces: compact iff sequentially compact (every sequence has a convergent subsequence) iff complete and totally bounded. Heine-Borel: $A\subseteq\mathbb{R}^n$ is compact iff closed and bounded. Continuous image of compact is compact; continuous real-valued function on compact set attains max and min. Hausdorff: compact subsets are closed; closed subsets of compact are compact.
Connectedness & Path-Connectedness
$X$ is connected if it cannot be written as a disjoint union of two non-empty open sets. $X$ is path-connected if for any $x,y\in X$, there exists a continuous path $\gamma:[0,1]\to X$ with $\gamma(0)=x$, $\gamma(1)=y$. Path-connected $\Rightarrow$ connected; converse fails (topologist's sine curve). Continuous image of connected is connected; this implies the intermediate value theorem. The connected components are the maximal connected subspaces.
Example 1
Prove $[0,1]$ is compact.
Solution: Let $\{U_\alpha\}$ be an open cover of $[0,1]$. Let $S=\{x\in[0,1]:$ the interval $[0,x]$ is covered by finitely many $U_\alpha\}$. $0\in S$ (trivially), $S$ is bounded above by 1, let $c=\sup S$. $c$ lies in some $U_\beta$, which contains $(c-\epsilon,c+\epsilon)$. Since $c-\epsilon/2\in S$ (as $c=\sup S$ and $c-\epsilon/2
Example 2
Show $\mathbb{R}$ is connected but not compact.
Solution: Connected: suppose $\mathbb{R}=U\cup V$, $U,V$ open, disjoint, non-empty. Pick $a\in U$, $b\in V$ (say $a
Practice
- Prove that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism.
- Show that the product of two compact spaces is compact (Tychonoff for finite products).
- Give an example of a connected but not path-connected space.
- Prove the intermediate value theorem using connectedness.
Show Answer Key
1. Let $f:X\to Y$ be a continuous bijection, $X$ compact, $Y$ Hausdorff. To show $f^{-1}$ is continuous: let $C\subseteq X$ be closed. Closed subset of compact is compact, so $C$ is compact. $f(C)$ is compact (continuous image). Compact in Hausdorff is closed. So $f(C)$ is closed. Thus $f$ maps closed sets to closed sets, i.e., $f^{-1}$ is continuous.
2. Let $X\times Y$ have an open cover $\{U_\alpha\}$. For each $x\in X$, $\{x\}\times Y$ is compact (homeomorphic to $Y$), covered by finitely many $U_\alpha$. Their union contains a tube $W_x\times Y$. The sets $\{W_x\}$ cover $X$; by compactness of $X$, finitely many suffice. The corresponding finite collection of $U_\alpha$'s covers $X\times Y$.
3. The topologist's sine curve: $S=\{(x,\sin(1/x)):x>0\}\cup\{(0,y):-1\le y\le 1\}$. It is connected (closure of the connected set $\{(x,\sin(1/x)):x>0\}$) but not path-connected (no path from $(0,0)$ to $(1/\pi,0)$ exists within $S$ — the oscillations prevent it).
4. Let $f:[a,b]\to\mathbb{R}$ be continuous with $f(a)