一般拓撲/緊開拓撲

定義（緊開拓撲）:

設 $X$ 和 $Y$ 為兩個拓撲空間，設 $F(X,Y)$ 為從 $X$ 到 $Y$ 的所有函式的集合。 $F(X,Y)$ 上的 **緊開拓撲** 被定義為：其子基由集合給出

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

其中 $K$ 在 $X$ 的所有緊子集上取值，而 $U$ 在 $Y$ 的所有開子集上取值。

命題（緊集上一致收斂的拓撲至少與連續函式空間上的緊開拓撲一樣細）:

設 $X$ 為一個拓撲空間，設 $Y$ 為一個一致空間。設 $S$ 為 $X$ 的所有緊子集的集合。 那麼 $C(X,Y)$ 上由 $S$-收斂的拓撲至少與 $C(X,Y)$ 上由緊開拓撲誘導的子空間拓撲一樣細。

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

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

構成 $K$ 的一個開覆蓋，因為每個 $E(x)$ 都是非空的，因此包含一個包含 $x$ 的開集。但 $K$ 是緊緻的，所以我們可以選擇一個有限子覆蓋 $U_{1},\ldots ,U_{n}$。根據定義，每個 $U_{j}$ 都與之前定義的 $U_{x}$ 之一相同，因此存在一個伴隨 $W_{j}$ 和一個點 $x_{j}\in X$ 使得 $f(U_{j})\subseteq W_{j}(f(x_{j}))$ 且 $(W_{j}\circ W_{j})(f(x_{j}))\subseteq U$。現在定義

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

我們斷言 $W(K,f)$ 是 $f$ 的一個鄰域，它包含在 $M(K,U)$ 內。事實上，設 $g\in W(K,f)$。如果 $y\in K$ 是任意的，則存在一個 $j\in \{1,\ldots ,n\}$ 使得 $y\in U_{j}$。從 $W(K,f)$ 的定義，我們推斷出 $(f(y),g(y))\in W\subseteq W_{j}$。然而，我們也知道 $(f(x_{j}),f(y))\in W_{j}$，因此 $(f(x_{j}),g(y))\in W_{j}\circ W_{j}\subseteq V_{j}$，由此得出 $g(y)\in V_{j}(f(x_{j}))\subseteq U$。由於 $y\in K$ 是任意的，因此 $g(K)\subseteq U$ 且 $g\in M(K,U)$。 $\Box$

命題（緊緻開拓撲和緊緻集上的一致收斂在區域性緊緻空間上的連續函式上重合）:

令 $X$ 是一個區域性緊空間，令 $Y$ 是一個一致空間。令 $S$ 是 $X$ 中所有緊集的集合。那麼由 $S$-收斂和緊開拓撲分別誘導的 $C(X,Y)$ 上的子空間拓撲是一致的。

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 $W(K,f)$ 內。事實上，假設 $g\in M(K_{1},W(f(x_{1})))\cap \cdots \cap M(K_{n},W(f(x_{n})))$，令 $x\in K$。令 $j\in \{1,\ldots ,n\}$ 使得 $x\in K_{j}$。由於 $g(K_{j})\subseteq W(f(x_{j}))$，特別是 $g(x)\in W(f(x_{j}))$。但是 $f(K_{j})\subseteq W(f(x_{j}))$ 也是，因此 $f(x)\in W(f(x_{j}))$。因此，

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

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