• Let $X$, $Y$ be topological spaces. If $K$ is a compact subspace of $X$ and $U$ is an open subset of $Y$, define $[K,U]=\{f\mid f\in C(X,Y) \text{ and } f(K)\subset U\}$. The sets $[K,U]$ form a subbasis for a topology on $C(X,Y)$ that is called the compact open topology.

  • A subbasis $S$ for a topology on $X$ is a collection of subsets of $X$ whose union equals $X$. The topology generated by the subbasis $S$ is defined to be the collection $T$ of all unions of finite intersections of elements of $S$.

So why the union of $[K,U]$s equals $C(X,Y)$?


Take $K = \emptyset, U = Y$. Then $[K, U] = C(X,Y)$.


There are two possible definitions of a subbasis: see Subbase in Wikipedia for a full discussion. The most frequently used definition does not require that the subbasis covers $X$. The trick is that finite intersection includes intersection indexed by the empty set, which by convention gives $X$.

  • $\begingroup$ I umderstand $[K_1, U_1]\cap[K_2, U_2]$, however, not be able to figure out what's $[K_1, U_1]\cup[K_2, U_2]$. $\endgroup$ – Brooks Feb 28 '16 at 14:31
  • 1
    $\begingroup$ Both are subsets of $C(X,Y)$, so this is just the usual intersection and usual union of two subsets. $\endgroup$ – J.-E. Pin Feb 28 '16 at 14:35

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