The Ascoli–Arzelà Theorem

By [L3], the compact subsets allowed in [L1] and [L2] are exactly the same subsets of $X$.

For every such $K$ and every open $V\subseteq Y$, both definitions use the identical subset $S(K,V)$ of $C(X,Y)$, including $K=\mathrm{\varnothing }$ and $V=Y$.

The two subbasic families are equal, and therefore generate equal topologies.

Suppose [L1] holds, and fix $x\in X$ and $\epsilon >0$. Choose its common neighbourhood $U$; by [L3], some ball $B(x,\delta )$ lies in $U$. The same $\delta$ works for every $f\in \mathcal{F}$, so [L2] holds.

Conversely suppose [L2] holds. For fixed $x$ and $\epsilon >0$, let $\delta >0$ be the common radius supplied there. The open neighbourhood $B(x,\delta )$ then satisfies [L1] for every $f\in \mathcal{F}$.

Steps 1.1 and 1.2 prove both directions without changing the order of the family quantifier.

If $X=\mathrm{\varnothing }$, the domain $C(X,Y)\times X$ is empty, so evaluation is continuous. Assume henceforth that $(f,x)\in C(X,Y)\times X$ and that $W\subseteq Y$ is open with $f(x)\in W$.

By [L4], $O=f^{-1}[W]$ is open and contains $x$. By [L2], choose open $U$ with $x\in U\subseteq K:=\bar{U}\subseteq O$ and $K$ compact.

The set $S(K,W)\times U$ is an open product neighbourhood of $(f,x)$: $f[K]\subseteq W$, $S(K,W)$ is subbasic open, and [L3] applies.

If $(g,y)\in S(K,W)\times U$, then $y\in U\subseteq K$ and $g(y)\in W$. Thus evaluation maps this neighbourhood into $W$, proving continuity.

Suppose $H$ is compact and fix $x\in X$. By [L1] and [L5], ${\pi }_x[H]$ is compact, and by [L4] and [L7] it is closed.

Conversely suppose $K_x:=\overline{\mathcal{F}(x)}$ is compact for every $x$. By [L2], $P:={\prod }_{x\in X}{K}_x$ is compact, including when $X=\mathrm{\varnothing }$, when it is a singleton.

Each $K_x$ is closed in the metric space $Y$ by [L4] and [L7], so $P={\bigcap }_{x\in X}{\pi }_x^{-1}[K_x]$ is closed in $Y^X$. Since $\mathcal{F}\subseteq P$, its closure $H$ is a closed subset of $P$.

Since $\mathcal{F}(x)\subseteq {\pi }_x[H]$, its closure is contained in ${\pi }_x[H]$; conversely continuity gives ${\pi }_x[H]\subseteq \overline{\mathcal{F}(x)}$. Hence $\overline{\mathcal{F}(x)}={\pi }_x[H]$ is compact.

By [L3], $H$ is compact. Together with steps 1.1--1.2 this proves both directions.

Fix $x\in X$ and $\epsilon >0$. By [L1], choose a neighbourhood $U$ of $x$ such that $d(f(y),f(x))<\epsilon \mathrm{/}3$ for every $f\in \mathcal{F}$ and $y\in U$.

Fix $h\in H$ and $y\in U$. The pointwise neighbourhood of $h$ requiring both $d(g(x),h(x))<\epsilon \mathrm{/}3$ and $d(g(y),h(y))<\epsilon \mathrm{/}3$ meets $\mathcal{F}$; choose $f$ in the intersection.

The triangle inequality and steps 1.1--1.2 give $d(h(y),h(x))<\epsilon$. The neighbourhood $U$ did not depend on $h$, so it proves equicontinuity of all of $H$.

For each fixed $h\in H$, the estimate in step 2.1 is the neighbourhood criterion for continuity at every $x$. Hence every member of $H$ is continuous.

For $x\in X$ and open $V\subseteq Y$, the pointwise subbasic set $\{g:g(x)\in V\}$ is $S(\{x\},V)$; the singleton is compact. Thus the compact-open topology on $\mathcal{E}$ is finer than the pointwise topology.

Fix $f\in \mathcal{E}\cap S(K,V)$. If $K=\mathrm{\varnothing }$ or $V=Y$, the whole family is a pointwise neighbourhood contained in $S(K,V)$. Otherwise, let $I$ be the set of triples $(x,r,U)$ with $x\in K$, $r>0$, $U$ an open neighbourhood of $x$, $B(f(x),3r)\subseteq V$, $d(f(y),f(x))<r$ on $U$, and $d(g(y),g(x))<r$ on $U$ for all $g\in \mathcal{E}$.

Openness of $V$, continuity of $f$, and [L2] show that every $x\in K$ occurs in some triple of $I$. Hence the open sets $U$ from all triples in $I$ cover $K$, without choosing one triple for every point.

Compactness of $K$ supplies finitely many triples $(x_i,r_i,U_i)$ from $I$ with $K\subseteq U_1\cup \cdots \cup U_m$. Let $N$ be the pointwise neighbourhood of $f$ in $\mathcal{E}$ defined by $d(g(x_i),f(x_i))<r_i$ for every $i$.

If $g\in N$ and $y\in K$, choose $i$ with $y\in U_i$. The three inequalities attached to $(x_i,r_i,U_i)$ and the definition of $N$ give $d(g(y),f(x_i))<3r_i$, so $g(y)\in V$. Hence $N\subseteq S(K,V)$.

Every compact-open subbasic neighbourhood has a pointwise neighbourhood inside it, so the pointwise topology on $\mathcal{E}$ is finer. Step 1.1 proves equality.

Let $H$ be the pointwise closure of $\mathcal{F}$ in $Y^X$. Pointwise relative compactness and [L1] make $H$ compact in the pointwise topology.

By [L2], $H\subseteq C(X,Y)$ and $H$ is equicontinuous. By [L3], its compact-open subspace topology equals its pointwise subspace topology, so $H$ is compact in the compact-open topology.

The family $\mathcal{F}$ is pointwise dense in $H$, and [L3] makes it compact-open dense there. Also, every pointwise subbasic set from [L5] is the compact-open set $S(\{x\},V)$ of [L4], so the compact-open topology is finer and the pointwise-closed set $H$ is compact-open closed. Thus the compact-open closure of $\mathcal{F}$ is exactly $H$, which is compact by step 2.1.

If $X=\mathrm{\varnothing }$ or $\mathcal{K}=\mathrm{\varnothing }$, the conclusion is vacuous. Otherwise fix $x\in X$ and $\epsilon >0$.

Call a pair $(O,U)$ of open sets admissible at $f\in \mathcal{K}$ when $f\in O$, $x\in U$, and $d(g(y),f(x))<\epsilon \mathrm{/}3$ for every $g\in O$ and every $y\in U$. Continuity of evaluation at $(f,x)$, which [L1] supplies, makes at least one pair admissible at each $f\in \mathcal{K}$. Let $\mathcal{A}$ be the set of all triples $(f,O,U)$ with $(O,U)$ admissible at $f$. This set is defined outright and no pair is selected, so no choice principle is used.

The open sets $O$ occurring in triples of $\mathcal{A}$ cover $\mathcal{K}$, because each $f\in \mathcal{K}$ lies in the $O$ of some admissible triple. Applying [L2] to the cover indexed by $\mathcal{A}$ yields finitely many triples $(f_1,O_1,U_1),\dots ,(f_m,O_m,U_m)$ of $\mathcal{A}$ whose $O_i$ already cover $\mathcal{K}$; only this finite selection is made. Put $U=U_1\cap \cdots \cap U_m$, an open neighbourhood of $x$ as a finite intersection.

Let $g\in \mathcal{K}$ and take $i\le m$ with $g\in O_i$. If $y\in U$ then $y\in U_i$ and $x\in U_i$, so admissibility at $f_i$ gives $d(g(y),f_i(x))<\epsilon \mathrm{/}3$ and $d(g(x),f_i(x))<\epsilon \mathrm{/}3$, whence $d(g(y),g(x))<2\epsilon \mathrm{/}3<\epsilon$. The one neighbourhood $U$ works for every $g\in \mathcal{K}$, which is what [L3] requires for equicontinuity at $x$.

If $X=\mathrm{\varnothing }$, [L2] is vacuous. Otherwise fix $x\in X$. For open $V\subseteq Y$, evaluation at $x$ pulls $V$ back to $S(\{x\},V)$, which is open by [L1]; hence evaluation is continuous.

By [L3], its image $\mathcal{K}(x)$ is compact. By [L4] and [L5], $\mathcal{K}(x)$ is closed.

Therefore $\overline{\mathcal{K}(x)}=\mathcal{K}(x)$ is compact. Since $x$ was arbitrary, [L2] proves pointwise relative compactness.

Suppose the compact-open closure ${\bar{\mathcal{F}}}^{\mathrm{c}\mathrm{o}}$ is compact. By [L2] it is equicontinuous, and restricting its common neighbourhood estimates to $\mathcal{F}$ shows that $\mathcal{F}$ is equicontinuous.

By [L3], ${\bar{\mathcal{F}}}^{\mathrm{c}\mathrm{o}}$ is pointwise relatively compact. For each $x$, the closure of $\mathcal{F}(x)$ is a closed subset of the compact closure of ${\bar{\mathcal{F}}}^{\mathrm{c}\mathrm{o}}(x)$, hence is compact by [L4]; thus $\mathcal{F}$ is pointwise relatively compact.

Conversely, if $\mathcal{F}$ is equicontinuous and pointwise relatively compact, [L1] says directly that its compact-open closure is compact.

Steps 1.1--1.2 and 1.3 prove the two directions of the equivalence.

By [L2], the compact space $X$ is locally compact; it is Hausdorff by hypothesis.

Apply [L1] to $X$, $Y$, and $\mathcal{F}$. This yields both directions of the claimed equivalence with the compact-open topology and the stated Choice hypothesis unchanged.

A uniform ball of radius $0<\delta <\min \{\epsilon ,1\}$ about $f$ is contained in every compact-convergence basic set $B_K(f,\epsilon )$, because its inequality holds at every point of $X$.

Conversely take $0<\epsilon <1$ and $g\in B_X(f,\epsilon )$, so $d(f(x),g(x))<\epsilon$ at every $x\in X$. Since $X$ is nonempty and compact, clause (U3) of [L2] makes $M:={\max }_{x\in X}d(f(x),g(x))$ exist, and $M<\epsilon <1$ because the maximum is one of the values. Hence $\bar{\rho }(f,g)={\sup }_{x\in X}\min \{d(f(x),g(x)),1\}=M<\epsilon$, so $g$ lies in the uniform ball of radius $\epsilon$. For the reverse inclusion note that step 1.1 is stated for a radius strictly below $\epsilon$ and so cannot be instantiated at $\epsilon$ itself; argue directly instead: if $\bar{\rho }(f,g)<\epsilon <1$ then $\min \{d(f(x),g(x)),1\}<\epsilon <1$ at every $x$, so $d(f(x),g(x))<\epsilon$ at every $x$ and $g\in B_X(f,\epsilon )$. So $B_X(f,\epsilon )$ is exactly that uniform ball.

Steps 1.1 and 2.1 show that compact convergence and uniform convergence induce the same topology. By [L1], that topology is also the compact-open topology.

By [L5] and [L6], $X$ with its metric topology is compact Hausdorff, so [L1] applies to $\mathcal{F}$.

By [L4] and [L2], the general compact-open topology used in [L1] is the uniform topology; hence the two closures and their compactness are identical.

By [L3], the equicontinuity condition in [L1] is exactly metric equicontinuity. Pointwise relative compactness is unchanged, so [L1] becomes the claimed equivalence in both directions.

Suppose $\mathcal{F}$ is compact in the uniform topology. Since this topology is metrizable, [L2] makes $\mathcal{F}$ closed; hence its uniform closure is itself.

Conversely suppose $\mathcal{F}$ is uniformly closed, equicontinuous, and pointwise bounded. For each $x$, the closure $\overline{\mathcal{F}(x)}$ is closed and remains bounded; this also holds when $\mathcal{F}(x)=\mathrm{\varnothing }$.

By [L1], $\mathcal{F}$ is equicontinuous and each $\overline{\mathcal{F}(x)}$ is compact. By [L2] each such coordinate closure is bounded, so $\mathcal{F}$ is pointwise bounded.

The coordinate closure $\overline{\mathcal{F}(x)}$ is closed and bounded, hence compact by properness of $Y$. Thus [L1] makes the uniform closure of $\mathcal{F}$ compact.

Since $\mathcal{F}$ is uniformly closed, it equals that compact closure and is compact. Steps 1.1--1.2 prove the converse implication, completing the equivalence.

For each $x\in X$, the coordinate set $\{f_k(x):k\in \mathbb{N}\}$ is bounded. Its closure is closed and remains bounded, and therefore is compact by properness.

By equicontinuity, step 1.1, and [L1], the uniform closure $H$ of $\{f_k:k\in \mathbb{N}\}$ is compact.

Choice implies the two weaker choice principles in [L2], so the compact metric space $H$ is sequentially compact. Hence $(f_k)$ has a subsequence converging in the uniform metric to some $f\in H\subseteq C(X,Y)$.

Convergence in the uniform metric is uniform convergence, which gives the claimed subsequence and continuous limit.

By [L3], both $\mathbb{R}$ and ${\mathbb{R}}^n$ are proper metric spaces: a closed bounded subset is compact.

By [L9] and [L10], $\bar{\rho }(f,g)=\min \{d_{\mathrm{\infty }}(f,g),1\}$, so balls of radius below $1$ agree and the supremum-metric topology is the uniform topology. In the real case, compactness then implies uniform closedness by [L4], and [L5] gives equicontinuity and pointwise boundedness. Conversely, uniform closedness plus those two conditions makes the compact closure supplied by [L5] equal to the family. The subsequence conclusion is [L6], and the stated uniform equicontinuity and common value bound are exactly [L7] and [L8].

For the Euclidean common bound, fix an equicontinuous, pointwise bounded family $\mathcal{F}\subseteq C(X,{\mathbb{R}}^n)$ and take the tolerance to be $1$. Equicontinuity of $\mathcal{F}$ gives, at every $x\in X$, a neighbourhood $U_x$ on which $\parallel f(y)-f(x){\parallel }_2<1$ for all $f\in \mathcal{F}$; Choice, which the Given supplies, licenses the family $(U_x{)}_{x\in X}$. Compactness of $X$ gives a finite subcover $U_{x_1},\dots ,U_{x_m}$.

Apply [L1] and [L2] to the proper target ${\mathbb{R}}^n$. This proves both numbered vector-valued assertions, including both directions of assertion 1.

Pointwise boundedness of the family $\mathcal{F}$ fixed in step 1.3 gives, for each of the finitely many $i\le m$, a number $M_i$ with $\parallel f(x_i){\parallel }_2\le M_i$ for every $f\in \mathcal{F}$. For $y\in X$ take $i$ with $y\in U_{x_i}$; then $\parallel f(y){\parallel }_2\le \parallel f(y)-f(x_i){\parallel }_2+\parallel f(x_i){\parallel }_2<1+M_i\le 1+{\max }_i{M}_i$. Thus one Euclidean bound serves all $f\in \mathcal{F}$ and all $y\in X$, and the converse from a common bound to pointwise boundedness is immediate.

For each $x\in X$, the closure $\overline{\mathcal{F}(x)}$ is a closed subset of the compact target $Y$, and is therefore compact by [L2]. This includes the empty coordinate set.

Thus $\mathcal{F}$ is pointwise relatively compact. Together with the assumed equicontinuity, [L1] makes its compact-open closure compact. No local compactness hypothesis on $X$ is used.

Every map $f:X\to Y$ is continuous because the inverse image of each open subset of $Y$ is a subset of $X$, hence open by [L1]. Thus $C(X,Y)=Y^X$.

The whole family $Y^X$ is equicontinuous: at $x\in X$, the neighbourhood $\{x\}$ makes $d(f(y),f(x))=0$ for every $f$ and every $y$ in it.

By [L4] and [L5], the pointwise topology on $Y^X$ is compact, including the empty product when $X=\mathrm{\varnothing }$.

By [L3], this pointwise topology equals the compact-open topology on the equicontinuous whole family. Hence $C(X,Y)$ is compact.

Define $\mathrm{\Phi }:Q\to C(X,Y)$ by $\mathrm{\Phi }(q)=c_q$. For a subbasic $S(K,V)$, its inverse image under $\mathrm{\Phi }$ is $Q$ when $K=\mathrm{\varnothing }$, and is $Q\cap V$ when $K\mathrm{\ne }\mathrm{\varnothing }$. Hence $\mathrm{\Phi }$ is continuous.

Fix $x_0\in X$. Evaluation at $x_0$ is continuous because the inverse image of open $V\subseteq Y$ is $S(\{x_0\},V)$. Its restriction to ${\mathcal{C}}_Q$ is inverse to $\mathrm{\Phi }$.

By [L2], the image ${\mathcal{C}}_Q=\mathrm{\Phi }[Q]$ is compact.

Therefore $\mathrm{\Phi }:Q\to {\mathcal{C}}_Q$ is a homeomorphism, and step 2.1 gives the asserted compactness.

The function $d$ satisfies the three axioms of [L3]: (M1) holds because $d(m,n)=0$ was defined to mean $m=n$; (M2) holds because the defining cases are symmetric in $m,n$; and for (M3), if $d(m,p)=0$ the inequality is trivial, while if $d(m,p)=1$ then $m\mathrm{\ne }p$, so $n$ differs from at least one of $m,p$ and the right side is at least $1$. So $d$ is a metric. Its metric topology is discrete because $B(n,1)=\{n\}$.

Every map $X\to Y$ is constant. The whole family $\mathcal{F}=C(X,Y)$ is equicontinuous, and $\mathcal{F}(\ast )=Y$ is bounded because it lies in the radius-$2$ ball about $0$.

Evaluation at $\ast$ is a bijection $\mathcal{F}\to Y$. By [L4], the inverse image of each open $V\subseteq Y$ is $S(\{\ast \},V)$, so evaluation is a homeomorphism for the compact-open topology.

The open cover $\{\{n\}:n\in \mathbb{N}\}$ of the infinite discrete space $Y$ has no finite subcover. Thus $Y$ and hence $\mathcal{F}$ are not compact, while the family is equicontinuous and pointwise bounded. Moreover $\overline{\mathcal{F}(\ast )}=Y$ is not compact, displaying exactly the missing hypothesis in [L1].