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I'm learning the proof of the following version of Arzela-Ascoli's theorem (Willard, General Topology, page 287):

Let $X$ be a Hausdorff, or regular, k-space, $(Y,\mathcal D)$ a Hausdorff uniform space, and $\mathcal F$ a family of continuous functions from $X$ to $Y$. Then $\mathcal F$ is compact in the compact-open topology iff

(a) $\mathcal F$ is pointwise closed,

(b) for each $x\in X$, $\pi_x(\mathcal F)$ has compact closure,

(c) $\mathcal F$ is equicontinuous on each compact subset of $X$.

($\mathcal D$ is a diagonal uniformity on $Y$. k-space := a topological space in which a set is closed iff its intersection with any compact set is closed. $f:X\to Y$ is said to be equicontinuous on $K\subseteq X$ iff $f|_K$ is equicontinuous.)

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Here is a sketch of Willard's proof:

Necessity:

$\mathcal F$'s compactness in the compact-open topology implies its compactness in the pointwise topology, which is (under Hausdorff assumption on $Y$) equivalent to (a) and (b).

(c), i.e. equicontinuity of $\mathcal F$ on every compact $K\subseteq X$, is shown by taking, for a given $x\in K$ and $E\in\mathcal D$, a symmetric $D\in\mathcal D$ with $D\circ D\subseteq E$ and finding (using regularity of $K$ and continuity of functions in $\mathcal F$) a finite cover of $\mathcal F_K:=\{f|_K:f\in\mathcal F\}$ of the form $$\left(U_{f_1}, D\left[f_1(x)\right]\right),\ldots,\left(U_{f_n}, D\left[f_n(x)\right]\right),$$ where $$(K, U):=\left\{f\in\mathcal F_K:f(K)\subseteq U\right\},\qquad D[y]:=\{z:(y,z)\in D\},$$ $f_i$ are elements of $\mathcal F_K$, and $U_{f_i}$ are closed neighbourhoods of $x$ in $K$ with $f_i(U_{f_i})\subseteq D[f_i(x)]$. Then $f(U_{f_1}\cap\ldots\cap U_{f_n})\subseteq E([f(x)])$ for all $f\in \mathcal F_K$.

Sufficiency can be shown like this (I think):

It is enough to show that (c) forces the compact-open topology on $\mathcal F$ to reduce to the pointwise topology (for we can then use the upper-mentioned equivalence of conditions (a)-(b) and compactness in the pointwise topology), i.e. to show that the pointwise convergence of nets in $\mathcal F$ implies their convergence in the compact-open topology on $\mathcal F$ = the topology of uniform convergence on $\mathcal F$ (as these two coincide on spaces of continuous functions).

Indeed, if a net $(f_\lambda)$ in $\mathcal F$ converges pointwise to $f\in\mathcal F$, then for every compact $K$ in $X$ $f_\lambda|_K\to f|_K$ pointwise, but, as on equicontinuous families the pointwise topology concides with the topology of uniform convergence on compacta, (c) implies that $f_\lambda|_K\to f|_K$ uniformly, so $f_\lambda$ converges to $f$ in the topology of uniform convergence on compacta.

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My question is: in which part of the proof is the assumption that $X$ is a k-space used? I think the property "$f:X\to Y$ is continuous iff its restriction on any compact $K\subset X$ is continuous" is meant to be used somewhere, but I don't see where. I should also mention that it is entirely possible that there is a hole in the above sketch of proof, as it is not word-by-word transcribed from Willard.

Thank you for your help!

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  • $\begingroup$ Hi, I was wondering if you came up with a solution to this problem. I'm reading Williard's book at the moment, and I was puzzled for the same reason you were. $\endgroup$ Sep 16, 2015 at 13:19

1 Answer 1

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You say $(f_{i})$ is a net in $\cal{F}$ converging pointwise to $f$. But $f$ does not belong necessarily to $\cal{F}$. So, the proof is going like this: If $K$ is compact, then $f_{i}|_{K}\rightarrow f|_{K}$ pointwise, so by equicontinuity of $\cal{F}$ on $K$, $f_{i}|_{K}\rightarrow f|_{K}$ in $\tau_{c}$, the compact-open topology. Thus, $f_{i}\rightarrow f$ in the topology of uniform convergence on compacta, by 43.6. Moreover, $(f_{i}|_{K})$ is an equicontinuous family, so $f|_{K}$ is continuous by 43.13 of Willard (: if $\cal{A}$ is equicontinuous family, the same holds for $\bar{\cal{A}}$). Thus, since $X$ is a k-space, $f$ is continuous. So, $\cal{F}\cup \{f\}$ is a family of continuous functions and by 43.7 (: for spaces of continuous functions the topology of compact convergence is the compact open topology $\tau_{c}$), $f_{i}\rightarrow f$ in $\tau_{c}$.

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