MATH 5345H — Week 13: Paracompactness; complete metric spaces

1 Complete Metric Spaces and Function Spaces

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Def. $(x_{n})$ in $(X,d)$ is Cauchy if $\forall\varepsilon>0\ \exists N$ : $d(x_{m},x_{n})<\varepsilon$ for all $m,n\geq N$ . $(X,d)$ is complete if every Cauchy sequence converges.
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Rmk. Every convergent sequence is Cauchy. The converse is the content of completeness.
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Lem. It suffices that every Cauchy sequence has a convergent subsequence.
Proof. Cauchy $+$ a subsequential limit $\Rightarrow$ the whole sequence converges to it.
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Thm. Every compact metric space is complete.
Proof. Sequential compactness supplies the subsequence.
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Thm. $\mathbb{R}^{n}$ is complete (in any norm-induced metric).

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Def. $(X,d)$ is totally bounded if for every $\varepsilon>0$ finitely many $\varepsilon$ -balls cover $X$ .
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Prop. Compact $\Rightarrow$ totally bounded.
Proof. The $\varepsilon$ -balls form an open cover.
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$\bigstar$ Thm. $(X,d)$ is compact $\Leftrightarrow$ complete and totally bounded.
Proof. $\Leftarrow$ : build a Cauchy subsequence. Cover by finitely many balls of radius $1$ ; one, $B_{1}$ , contains $x_{n}$ for $n$ in an infinite set $J_{1}$ . Inductively pick $B_{k+1}$ of radius $1/(k+1)$ containing $x_{n}$ for infinitely many $n\in J_{k}$ , giving $J_{1}\supset J_{2}\supset\cdots$ . Choose $n_{1}<n_{2}<\cdots$ with $n_{k}\in J_{k}$ ; then $x_{n_{i}},x_{n_{j}}\in B_{k}$ for $i,j\geq k$ , so the subsequence is Cauchy. Complete $\Rightarrow$ convergent $\Rightarrow$ sequentially compact $\Rightarrow$ compact (§28).
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Caution. Total boundedness alone is not enough — and boundedness alone is far weaker than total boundedness in general metric spaces.