2.7 Nets and compactness796552E62FD2

Proof. Suppose that $X$ is compact, and let $(x_ i)_{i\in I}$ be a net in $X$. The closures $T_{i_0}$ of the tails $\{ x_ i:i\geq i_0\} $ have the finite intersection property, since finitely many of them contain the tail of a common upper bound of their indices. As $X$ is compact, they have a common point $x$; otherwise their complements would form an open cover without a finite subcover. Thus every neighborhood of $x$ meets every tail of the net.

We now construct a subnet converging to $x$. Let $J$ be the set of triples $(i,U,k)$ with $i,k\in I$, $k\geq i$, $U$ an open neighborhood of $x$ and $x_ k\in U$, preordered by

It contains $(i,X,i)$ for every $i$, and it is directed: given two triples $(i_1,U_1,k_1)$ and $(i_2,U_2,k_2)$, choose $i\geq i_1,i_2,k_1,k_2$, put $U=U_1\cap U_2$, and choose $k\geq i$ with $x_ k\in U$, which is possible because $U$ meets the tail of $i$; then $(i,U,k)$ dominates both triples. The map $h(i,U,k)=k$ is increasing, and it is final because $(i,U,k)\geq (i_0,X,i_0)$ implies $k\geq i_0$. The subnet $(x_{h(j)})_{j\in J}$ converges to $x$: if $V$ is a neighborhood of $x$, choose an open neighborhood $U\subset V$ of $x$ and a triple $(i,U,k)\in J$; then $x_{k'}\in U'\subset V$ for every $(i',U',k')\geq (i,U,k)$.

Conversely, suppose that $X$ is not compact, and let $\mathcal U$ be an open cover without a finite subcover. The finite subfamilies $F\subset \mathcal U$, ordered by inclusion, form a directed set, and we may choose points $x_ F\notin \bigcup F$. If a subnet $(x_{h(j)})_{j\in J}$ converged to a point $x$, let $U\in \mathcal U$ contain $x$. Then $x_{h(j)}\in U$ for all large $j$, while finality gives $h(j)\supset \{ U\} $, and hence $x_{h(j)}\notin U$, for all large $j$. Since $J$ is directed, some $j$ satisfies both conditions, a contradiction. $\square$

Proof. This is Tychonoff's theorem; we prove it with ultrafilters, using the axiom of choice in the form of Zorn's lemma. A filter on a set $Z$ is a family $\mathcal F$ of subsets of $Z$ such that $Z\in \mathcal F$, $\varnothing \notin \mathcal F$, and $\mathcal F$ is closed under finite intersections and under passing to supersets. An ultrafilter is a filter which is maximal for inclusion. Since the union of a chain of filters containing a given filter is again a filter, Zorn's lemma shows that every filter is contained in an ultrafilter.

A filter $\mathcal U$ on $Z$ is an ultrafilter if and only if, for every $A\subset Z$, either $A\in \mathcal U$ or $Z\setminus A\in \mathcal U$. Indeed, let $\mathcal U$ be an ultrafilter and $A\subset Z$. If $A$ meets every member of $\mathcal U$, the sets containing some $U\cap A$ with $U\in \mathcal U$ form a filter containing $\mathcal U$ and $A$, so $A\in \mathcal U$ by maximality; otherwise some $U\in \mathcal U$ is disjoint from $A$, and $Z\setminus A\supset U$ belongs to $\mathcal U$. Conversely, if $\mathcal U$ has this property, a strictly larger filter would contain some $A\notin \mathcal U$ together with $Z\setminus A$, hence $\varnothing $.

A filter on a topological space $Z$ converges to $z$ if every neighborhood of $z$ belongs to it. We show that $Z$ is compact if and only if every ultrafilter on $Z$ converges. If $Z$ is compact and $\mathcal U$ is an ultrafilter, the closures of the members of $\mathcal U$ have the finite intersection property and hence a common point $z$. If an open neighborhood $N$ of $z$ were not in $\mathcal U$, the closed set $Z\setminus N$ would belong to $\mathcal U$. Since $z$ lies in the closure of every member of $\mathcal U$, this would give $z\in Z\setminus N$, which is absurd. Hence $\mathcal U$ converges to $z$. Conversely, suppose that every ultrafilter converges, and let $\mathcal C$ be a family of closed sets with the finite intersection property. The sets containing a finite intersection of members of $\mathcal C$ form a filter, which lies in an ultrafilter $\mathcal U$ converging to some $z$. If $z\notin C$ for some $C\in \mathcal C$, then both $C$ and its complement, a neighborhood of $z$, would belong to $\mathcal U$. Hence $z$ lies in every member of $\mathcal C$, and $Z$ is compact.

Now let $Z=\prod _\alpha X_\alpha $ with compact factors and projections $p_\alpha $; if some $X_\alpha $ is empty, so is $Z$, and there is nothing to prove. Let $\mathcal U$ be an ultrafilter on $Z$. For each $\alpha $, the family $\{ A\subset X_\alpha :p_\alpha ^{-1}(A)\in \mathcal U\} $ is a filter with the property characterizing ultrafilters, hence an ultrafilter, and since $X_\alpha $ is compact it converges to some $z_\alpha \in X_\alpha $. Put $z=(z_\alpha )$. A basic neighborhood $\bigcap _{l=1}^ np_{\alpha _ l}^{-1}(V_ l)$ of $z$, with $V_ l$ a neighborhood of $z_{\alpha _ l}$, is a finite intersection of members of $\mathcal U$, so every neighborhood of $z$ belongs to $\mathcal U$. Thus every ultrafilter on $Z$ converges, and $Z$ is compact. $\square$

Proof. We embed $\mathcal M$ homeomorphically onto a closed subset of a product of compact intervals. For $f\in C_ c(\Omega ,\mathbb R)$, choose $r(f)$ with $\operatorname{Supp}f\subset K_{r(f)}$; this is possible because $\operatorname{Supp}f$ is compact and the open sets $\operatorname{Int}K_ r$ increase to $\Omega $. For $\mu \in \mathcal M$ we have $|f|\leq \| f\| _\infty \chi _{r(f)}$, so $\int f\, \mathrm d\mu $ lies in $I_ f=[-M_{r(f)}\| f\| _\infty ,M_{r(f)}\| f\| _\infty ]$. The product $P=\prod _ fI_ f$ over all $f\in C_ c(\Omega ,\mathbb R)$ is compact by lemma 2.7.2. The map $\Phi \colon \mathcal M\to P$, $\Phi (\mu )=(\int f\, \mathrm d\mu )_ f$, is injective by the uniqueness in theorem 2.3.3, and it is a homeomorphism onto its image. Indeed, since $\int f\, \mathrm d\mu =\int \operatorname{Re}f\, \mathrm d\mu +\mathrm i\int \operatorname{Im}f\, \mathrm d\mu $, the vague topology on $\mathcal M$ is the weakest topology for which the maps $\mu \mapsto \int f\, \mathrm d\mu $ with real $f$ are continuous. The subspace topology of $\Phi (\mathcal M)$ is the weakest topology for which the restrictions of the coordinate projections are continuous, and $\Phi $ intertwines these two families of maps.

The image consists of the points $(a_ f)\in P$ such that $a_{sf+tg}=sa_ f+ta_ g$ for all $f,g$ and real $s,t$, $a_ f\geq 0$ whenever $f\geq 0$, and $a_{\chi _ r}\leq M_ r$ for every $r$: a point satisfying these conditions defines a positive linear functional on $C_ c(\Omega ,\mathbb R)$, which by theorem 2.3.3 is integration against a Radon measure belonging to $\mathcal M$. Each condition involves finitely many coordinates and is closed, so $\Phi (\mathcal M)$ is closed in $P$ and hence compact. Therefore $\mathcal M$ is vaguely compact, and the assertion about nets follows from proposition 2.7.1. $\square$

Proof. We may assume that $K\neq \varnothing $. Let $\varepsilon >0$. By equicontinuity, every $x\in K$ is the center of an open ball on which $|f-f(x)|<\varepsilon $ for all $f\in \mathcal F$; by compactness, $K$ is covered by finitely many such balls $B_ l$ with centers $x_ l$, $1\leq l\leq N$. Then every $f\in \mathcal F$ is bounded on $K$ by $M=\varepsilon +\max _ l\sup _{g\in \mathcal F}|g(x_ l)|$, which is finite by pointwise boundedness. Let $G$ be a finite set of scalars such that every scalar of absolute value at most $M$ lies at distance less than $\varepsilon $ from some element of $G$. For each map $\gamma \colon \{ 1,\ldots ,N\} \to G$, the functions $f\in \mathcal F$ with $|f(x_ l)-\gamma (l)|<\varepsilon $ for all $l$ form a set $\mathcal F_\gamma $, and these finitely many sets cover $\mathcal F$. If $f,g\in \mathcal F_\gamma $ and $y\in B_ l$, then

Choosing one function in each nonempty $\mathcal F_\gamma $, we obtain a finite set within uniform distance $4\varepsilon $ of $\mathcal F$, and hence within $5\varepsilon $ of its closure $\overline{\mathcal F}$ in $C(K)$. Thus $\overline{\mathcal F}$ is totally bounded. It is also complete, because $C(K)$ is complete for the uniform metric: a uniform Cauchy sequence converges pointwise, and the limit is continuous since the convergence is uniform.

A complete and totally bounded metric space $Y$ is sequentially compact and compact. Given a sequence in $Y$, cover $Y$ by finitely many sets of diameter less than $1$ and keep one containing infinitely many terms. Cover it by finitely many sets of diameter less than $1/2$, keep one containing infinitely many of those terms, and so on. A diagonal choice gives a Cauchy subsequence, which converges because $Y$ is complete.

Suppose now that an open cover of $Y$ had no finite subcover, and put $S_0=Y$. Covering $S_{n-1}$ by finitely many subsets of diameter less than $1/n$, we see that one of them, $S_ n$, is not covered by finitely many members of the cover, since otherwise $S_{n-1}$ would be. This gives nested nonempty sets $S_1\supset S_2\supset \cdots $ with $\operatorname {diam}S_ n<1/n$, none of which is covered by finitely many members of the cover. Points $y_ n\in S_ n$ form a Cauchy sequence with a limit $y$. If the open ball of radius $s$ about $y$ in $Y$ lies in a member $U$ of the cover, then $S_ n$ lies in this ball, and hence in $U$, as soon as $1/n+\operatorname {dist}(y_ n,y)<s$, a contradiction. Hence $\overline{\mathcal F}$ is compact, and every sequence in $\mathcal F$ has a uniformly convergent subsequence. $\square$