The Ascoli–Arzelà Theorem: Examples and Counterexamples

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].

The map $x\mapsto 1-\mathrm{\mid }x-k\mathrm{\mid }$ is $1$-Lipschitz, and taking the maximum with $0$ preserves that bound. Hence every $f_k$ is continuous and $1$-Lipschitz.

Let $K\subseteq \mathbb{R}$ be compact. If $K=\mathrm{\varnothing }$, convergence on $K$ is vacuous. Otherwise [L3] gives $R>0$ with $\mathrm{\mid }x\mathrm{\mid }\le R$ for $x\in K$, and [L4] gives $N\in \mathbb{N}$ with $N>R+1$.

If $k\ge N$ and $x\in K$, then $\mathrm{\mid }x-k\mathrm{\mid }\ge k-\mathrm{\mid }x\mathrm{\mid }>1$, so $f_k(x)=0$. Thus the sequence is eventually identically zero on every compact $K$, and therefore converges to zero uniformly on each compact set.

By [L1], [L2], and [L6], step 2.1 is convergence in the general compact-open topology.

Yet $f_k(k)=1$ for every $k$, so no tail has $\mathrm{\mid }{f}_k(x)\mathrm{\mid }<1\mathrm{/}2$ for every $x\in \mathbb{R}$. By [L5], convergence is not uniform on the whole real line.

By [L2], $P$ is compact. Define $\mathrm{\Phi }:P\to C([0,1],\mathbb{R})$ by $\mathrm{\Phi }(a,b)=f_{a,b}$.

For $0\le t\le 1$,

Put $M=\max \{\mathrm{\mid }\alpha \mathrm{\mid },\mathrm{\mid }\beta \mathrm{\mid },\mathrm{\mid }\gamma \mathrm{\mid },\mathrm{\mid }\delta \mathrm{\mid }\}$. Every member satisfies $\mathrm{\mid }{f}_{a,b}(s)-f_{a,b}(t)\mathrm{\mid }=\mathrm{\mid }b-a\mathrm{\mid }\mathrm{\mid }s-t\mathrm{\mid }\le 2M\mathrm{\mid }s-t\mathrm{\mid }$, so the family is equicontinuous, including the degenerate case $M=0$.

By [L1], $\mathcal{A}=\mathrm{\Phi }[P]$ is compact.

The endpoint map $E:\mathcal{A}\to P$, $E(f)=(f(0),f(1))$, is continuous because uniform distance controls both endpoint differences, and $E\circ \mathrm{\Phi }$ and $\mathrm{\Phi }\circ E$ are identity maps. Hence $\mathrm{\Phi }$ is a homeomorphism onto $\mathcal{A}$.

For fixed $t$, the coordinate set $\{f_{a,b}(t):(a,b)\in P\}$ is the continuous image of compact $P$, hence compact by [L1]. It is therefore already its compact closure, proving pointwise relative compactness.