Page 246 - DMTH503_TOPOLOGY

Topology

Notes If C is compact, this set is open in C because f|C is continuous. Since X is compactly generated,
it follows that f (V) is open in X.
–1
Theorem 1: Let X be a compactly generated space. Let (Y, d) be a metric space. Then (X, Y) is
X
closed in Y in the topology of compact convergence.
X
Proof: Let f  Y be a limit point of (X, Y); we wish to show f is continuous. It suffices to show
that f |C is continuous for each compact subspace C of X. For each n consider the neighbourhood
1
B (f, 1/n) of f; it intersects (X, Y), so we can choose a function f  C(X, Y) lying in this
c n
neighbourhood. The sequence of functions f | C : C  Y converges uniformly to the function
n
f|C, so that by the uniform limit theorem, f | C is continuous.
29.1.4 Compact-open Topology

Definition: Let X and Y be topological spaces. If C is a compact subspace of X and U is an open
subset of Y, define
S(C, U) = {f | f (X, Y) and f(C)  U}
The sets S(C, U) form a sub-basis for a topology on (X, Y) that is called the compact-open
topology.

Theorem 2: Let X be a space and let (Y, d) be a metric space. On the set (X, Y), the compact-open
topology and the topology of compact convergence coincide.
Proof: If A is a subset of Y and  > 0, let U(A, ) be the  - neighbourhood of A. If A is compact
and V is an open set containing A, then there is an  > 0 such that U(A, ). Indeed, the minimum
value of the function d(a, X – V) is the required .

We first prove that the topology of compact convergence is finer than the compact-open topology.
Let S(C, U) be a sub-basis element for the compact-open topology, and let f be an element of
S(C, U). Because f is continuous, f(C) is a compact subset of the open set U. Therefore, we can
choose  so that  - neighbourhood of f(C) lies in U. Then, as desired.
B (f, )  S(C, U)
C
Now we prove that the compact-open topology is finer than the topology of compact convergence.
Let f  (X, Y). Given an open set about f in the topology of compact convergence, it contains a
basis element of the form B (f, ). We shall find a basis element for the compact-open topology
C
that contains f and lies in B (f, ).
C
Each point x of X has a neighbourhood V such that F(V ) lies in an open set U of Y having
x x x
diameter less than . [For example, choose V so that f(V ) lies in the /4-neighbourhood of f(x).
x x
Then f(V ) lies in the /3-neighbourhood of f(x), which has diameter at most 2/3]. Cover C by
x
finitely many such sets V , say for x = x ,...,x . Let C = V C. Then C is compact, and the basis
x 1 n x x x
element.
S(C ,U ) ... S(C ,U )
1 x x 1 n x n x
Theorem 3: Let X be locally compact Hausdorff; let e (X, Y) have the compact-open topology.
Then the map
e : X × e (X, Y)  Y
defined by the equation
e (x, f) = f (x)
is continuous.

The map e is called the evaluation map.

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