0.1  (§27) Compact subspaces of the real line

• $\mathrm{★}$ Thm. Every closed interval $[a,b]\subset ℝ$ is compact.

  Proof. Let $S=\{x\in [a,b]\mid [a,x]\text{ has a finite subcover}\}$, $c=supS$. Show $c\in S$ (an open $U_{\beta }\ni c$ reaches back past some $x\in S$), then $c=b$ (otherwise $U_{\beta }$ reaches past $c$). Least upper bound property again.

• $\mathrm{★}$ Thm. $Y\subset {ℝ}^n$ is compact $\iff$ $Y$ is closed and bounded.

  Proof. $\Leftarrow$: $Y$ is closed inside some $\prod [a_i,b_i]$. $\Rightarrow$: cover by the cubes ${(-M,M)}^n$; and ${ℝ}^n$ Hausdorff gives closed.

• Ex. $(a,b)$, $[a,b)$, $(a,b]$ are not closed in $ℝ$, hence not compact.

• Ex. $f:[0,1]\to S^1$ is a quotient map ($[0,1]$ compact, $S^1$ Hausdorff); likewise $f\times f:{[0,1]}^2\to T^2$.

• $\mathrm{★}$ Thm. (Extreme value theorem.) $f:X\to ℝ$ continuous, $X$ compact $\Rightarrow$ there are $c,d\in X$ with $f(c)\le f(x)\le f(d)$ for all $x$.

  Proof. $f(X)$ compact $\Rightarrow$ closed and bounded $\Rightarrow$ contains its inf and sup.

• Def. $d(x,A)=inf\{d(x,a)\mid a\in A\}$; $\mathrm{diam}(A)=sup\{d(a,b)\mid a,b\in A\}$.

• Lem. $x\mapsto d(x,A)$ is continuous; indeed $|d(x,A)-d(y,A)|\le d(x,y)$.

• $\mathrm{★}$ Lem. (Lebesgue number lemma.) $𝒜$ an open cover of a compact metric space $(X,d)$ $\Rightarrow$ there is $\delta >0$ such that every $B\subset X$ with lies in a single element of $𝒜$.

  Proof. Take a finite subcover $U_1,\mathrm{\dots },U_n$, put $C_i=X-U_i$ and $f(x)=\frac{1}{n}{\sum }_{i}d(x,C_i)$. Then $f>0$; let $\delta =\min f>0$ by the extreme value theorem.

• Def. $f$ is uniformly continuous if $\forall \epsilon >0\exists \delta >0$: — one $\delta$ for all points.

• $\mathrm{★}$ Thm. $f:X\to Y$ continuous, $X$ compact metric $\Rightarrow$ $f$ is uniformly continuous.

  Proof. Take a Lebesgue number for ${\{f^{-1}(B(y,\epsilon /2))\}}_{y\in Y}$.

0.2  (§28) Limit point compactness

• Def. $X$ is limit point compact if every infinite subset has a limit point.

• Thm. Compact $\Rightarrow$ limit point compact.

  Proof. If $A$ has no limit point, $A$ is closed and each $p\in A$ has $U_p$ with $A\cap U_p=\{p\}$. Cover by $X-A$ and the $U_p$; finiteness forces $A$ finite.

• Def. A subsequence $x_{n_1},x_{n_2},\mathrm{\dots }$ with . $X$ is sequentially compact if every sequence has a convergent subsequence.

• $\mathrm{★}$ Thm. For metrizable $X$, TFAE: (1) compact; (2) limit point compact; (3) sequentially compact.

  Proof. $(1)\Rightarrow (2)$ above. $(2)\Rightarrow (3)$: if $A=\{x_n\}$ is finite use a constant subsequence; else take a limit point $p$ and pick $x_{n_k}\in A\cap B(p,1/k)$ inductively. $(3)\Rightarrow (1)$ needs two lemmas below.

• Lem. $X$ sequentially compact metric $\Rightarrow$ the Lebesgue number lemma holds for $X$.

  Proof. Else get $C_n$ of diameter in no element of $𝒜$; a convergent subsequence $x_{n_k}\to p$ traps $C_{n_k}$ inside a ball around $p$ — contradiction.

• Lem. $X$ sequentially compact metric $\Rightarrow$ $X$ is totally bounded (finitely many $\epsilon$-balls cover $X$, each $\epsilon >0$).

  Proof. Else build $x_n$ with mutual distances $\ge \epsilon$; no convergent subsequence.

• Rmk. Then $(3)\Rightarrow (1)$: given $𝒜$ with Lebesgue number $\delta$, cover by $\epsilon$-balls with $\epsilon =\delta /3$; each has diameter , so each sits in one element of $𝒜$.