Page 245 - DMTH503_TOPOLOGY

Unit 29: Pointwise and Compact Convergence

I
This example shows that the subspace (I, ) of continuous functions is not closed in in the Notes
topology of pointwise convergence.
29.1.2 Compact Convergence

X
Definition: Let (Y, d) be a metric space; let X be a topological space. Given an element f of Y , a
compact subspace C of X, and a number  > 0, let B (f, ) denote the set of all those elements g
C
of Y for which
X
sup {d(f(x), g(x))|x  C} < 
The sets B (f, ) form a basis for a topology on Y . It is called the topology of compact convergence
X
C
(or sometimes the “topology of uniform convergence on compact sets”).
It is easy to show that the sets B (f, ) satisfy the conditions for a basis. The crucial step is to note
C
that if g  B (f, ), then for
C
 =  – sup{d(f(x), g(x))|x  C},
we have B (g, )  B (f, )
C C

Note The topology of compact convergence differs from the topology of pointwise
convergence in that the general basis element containing f consists of functions that are
“close” to f not just at finitely many points, but at all points of some compact set.

29.1.3 Compactly Generated

Definition: A space X is said to be compactly generated if it satisfies the following condition. A
set A is open in X if A C is open in C for each compact subspace C of X.
This condition is equivalent to requiring that a set B be closed in X if B C is closed in C for each
compact C. It is a fairly mild restriction on the space; many familiar spaces are compactly
generated.
Lemma 1: If X is locally compact, or if X satisfies the first countability axiom, then X is compactly
generated.
Proof: Suppose that X is locally compact. Let A C be open in C for every compact subspace C
of X. We show A is open in X. Given x  A, choose a neighbourhood U of x that lies in a compact
subspace C of X. Since A C is open in C by hypothesis, A U is open in U, and hence open
in X. Then A U is a neighbourhood of x contained in A, so that A is open in X.
Suppose that X satisfies the first countability axiom. If B C is closed in C for each compact
subspace C of X, we show that B is closed in X. Let x be a point of B; we show that x  B. Since
X has a countable basis at x, there is a sequence (x ) of points of B converging to x. The subspace
n
C = {x} {x |n  ]
n +
is compact, so that B C is by assumption closed in C. Since B C contains x for every n, it
n
contains x as well. Therefore, x  B, as desired.
Lemma 2: If X is compactly generated, then a function f : X  Y is continuous if for each compact
subspace C of X, the restricted function f|C is continuous.
Proof: Let V be an open subset of Y; we show that f (V) is open in X. Given any subspace C of X.
–1
–1
f (V) C = (f | C) (V)
–1

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