一般拓扑/紧开拓扑

定义（紧开拓扑）:

设 $X$ 和 $Y$ 为两个拓扑空间，设 $F(X,Y)$ 为从 $X$ 到 $Y$ 的所有函数的集合。 $F(X,Y)$ 上的 **紧开拓扑** 被定义为：其子基由集合给出

$M(K,U):=\{f\in F(X,Y)|f(K)\subseteq U\}$,

其中 $K$ 在 $X$ 的所有紧子集上取值，而 $U$ 在 $Y$ 的所有开子集上取值。

命题（紧集上一致收敛的拓扑至少与连续函数空间上的紧开拓扑一样细）:

设 $X$ 为一个拓扑空间，设 $Y$ 为一个一致空间。设 $S$ 为 $X$ 的所有紧子集的集合。 那么 $C(X,Y)$ 上由 $S$-收敛的拓扑至少与 $C(X,Y)$ 上由紧开拓扑诱导的子空间拓扑一样细。

Proof: We prove that any neighbourhood of an arbitrary $f\in C(X,Y)$ in the compact-open topology contains a neighbourhood of $f$ in the topology of uniform convergence on compact subsets of $X$. Thus, let $f\in C(X,Y)$ be arbitrary. Thus, suppose that $f\in M(K,U)$, where $K\subseteq X$ is compact and non-empty and $U\subseteq Y$ is open; any neighbourhood of $f$ with respect to the compact-open topology will be the finite intersection of sets of this form. Let now $x\in K$ be arbitrary. By the definition of the topology induced by a uniform space, the set of those entourages $V_{x}$ of $Y$ such that $V_{x}(f(x))\subseteq U$ is nonempty. Moreover, for each such $V_{x}$, we may choose an entourage $W_{x}$ of $Y$ such that $W_{x}\circ W_{x}\subseteq V_{x}$. For each such entourage, let $U_{x}$ be an open neighbourhood of $x$ such that $f(U_{x})\subseteq W_{x}(f(x))$. We shall denote the collection of all such $U_{x}$ by $E(x)$. Then the union of all these $E(x)$, ie. the collection

$\bigcup _{x\in K}E(x)$,

构成 $K$ 的一个开覆盖，因为每个 $E(x)$ 都是非空的，因此包含一个包含 $x$ 的开集。但 $K$ 是紧致的，所以我们可以选择一个有限子覆盖 $U_{1},\ldots ,U_{n}$。根据定义，每个 $U_{j}$ 都与之前定义的 $U_{x}$ 之一相同，因此存在一个伴随 $W_{j}$ 和一个点 $x_{j}\in X$ 使得 $f(U_{j})\subseteq W_{j}(f(x_{j}))$ 且 $(W_{j}\circ W_{j})(f(x_{j}))\subseteq U$。现在定义

$W:=W_{1}\cap \cdots \cap W_{n}$.

我们断言 $W(K,f)$ 是 $f$ 的一个邻域，它包含在 $M(K,U)$ 内。事实上，设 $g\in W(K,f)$。如果 $y\in K$ 是任意的，则存在一个 $j\in \{1,\ldots ,n\}$ 使得 $y\in U_{j}$。从 $W(K,f)$ 的定义，我们推断出 $(f(y),g(y))\in W\subseteq W_{j}$。然而，我们也知道 $(f(x_{j}),f(y))\in W_{j}$，因此 $(f(x_{j}),g(y))\in W_{j}\circ W_{j}\subseteq V_{j}$，由此得出 $g(y)\in V_{j}(f(x_{j}))\subseteq U$。由于 $y\in K$ 是任意的，因此 $g(K)\subseteq U$ 且 $g\in M(K,U)$。 $\Box$

命题（紧致开拓扑和紧致集上的一致收敛在局部紧致空间上的连续函数上重合）:

令 $X$ 是一个局部紧空间，令 $Y$ 是一个一致空间。令 $S$ 是 $X$ 中所有紧集的集合。那么由 $S$-收敛和紧开拓扑分别诱导的 $C(X,Y)$ 上的子空间拓扑是一致的。

Proof: We prove that both topologies generate the same neighbourhood systems. In view of the fact that the topology of uniform convergence on compact sets on spaces of continuous functions is at least as fine as the compact-open topology, it is sufficient to show that any neighbourhood of an arbitrary $f\in C(X,Y)$ with respect to the topology of uniform convergence on compact subsets contains a neighbourhood of $f$ with respect to the compact-open topology. Hence, let $V$ be any entourage of $Y$ and let $K\subseteq X$ be compact, so that $V(K,f)$ represents an arbitrary element of the canonical neighbourhood basis of $f$ with respect to the topology of uniform convergence on compact sets. We choose an entourage $W$ of $Y$ such that $W\circ W\subseteq V$. Now $X$ is locally compact, so that for each point $x\in K$, the collection of compact neighbourhoods $K_{x}$ of $x$ such that $f(K_{x})\subseteq W(f(x))$ is non-empty. The collection of all those $K_{x}$ we shall denote by $E(x)$. Now the collection of all ${\overset {\circ }{K_{x}}}$ (where $x$ ranges over all of $K$) is an open cover of $K$, whence we may choose a finite subcover ${\overset {\circ }{K_{1}}},\ldots ,{\overset {\circ }{K_{n}}}$. Since the interior is a subset of its original set, the sets $K_{1},\ldots ,K_{n}$ cover $K$. Moreover, by definition, each $K_{j}$ has an $x_{j}\in K_{j}$ such that $f(K_{j})\subseteq W(f(x_{j}))$. We claim that

$M(K_{1},W(f(x_{1})))\cap \cdots \cap M(K_{n},W(f(x_{n})))$

包含在 $W(K,f)$ 内。事实上，假设 $g\in M(K_{1},W(f(x_{1})))\cap \cdots \cap M(K_{n},W(f(x_{n})))$，令 $x\in K$。令 $j\in \{1,\ldots ,n\}$ 使得 $x\in K_{j}$。由于 $g(K_{j})\subseteq W(f(x_{j}))$，特别是 $g(x)\in W(f(x_{j}))$。但是 $f(K_{j})\subseteq W(f(x_{j}))$ 也是，因此 $f(x)\in W(f(x_{j}))$。因此，

$(f(x_{j}),f(x))\in W\wedge (f(x_{j}),g(x))\in W\Rightarrow (f(x),g(x))\in V$,

由于$x\in K$ 是任意的，所以$g\in V(K,f)$. $\Box$