Page 145 - DMTH503_TOPOLOGY

Unit 15: Local Compactness

Theorem 1: Let (X, T) and (Y, ) be topological spaces and f : (X, T) onto (Y, U) be a continuous open Notes

map. Then if X is locally compact, then Y is also.
Or
Every open continuous image of a locally compact space is locally compact.
Proof: Let f : (X, T)  (Y, U) be a continuous open map and X a locally compact space.
We claim Y is locally compact.
Let y  Y be arbitrary and U  Y a nbd of y.
y  Y, f : X  Y is onto   x  X s.t. f(x) = y

f is continuous
 Given any nbd U of y,  a nbd V  X of x s.t. f(V)  U. X is locally compact.
 X is locally compact at x and V is a nbd of x.
  is compact set A s.t. x  Aº  A  V
 f(x)  f(Aº)  f(A)  f(V)  U

 y  f(Aº)  f(A)  U ...(1)
Now, f is open, Aº  X is open.
 f(Aº)  Y is open
 f(Aº) = [f(Aº)]º ...(2)
From (1), f(Aº)  f(A)
Thus [f(Aº)]º  [f(A)]º

 f(Aº)  [f(A)]º, (on using (2))
 f(Aº)  [f(A)]º  f(A)
Using this in (1),
y  f(Aº)  [f(A)]º  f(A)  U
or y  [f(A)]º  f(A)  U

Taking B = f(A) = continuous image of compact set A
= compact set
We obtain y  Bº  B  U, B is compact.
Finally, we have shown that given any y  Y and a nbd U of y,  a compact set B  Y, s.t.
y  Bº  B  U.
Hence Y is locally compact at y so that Y is locally compact.
Theorem 2: Every locally compact T -space is a regular space.
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Proof: Let (X, T) be a locally compact T -space. To prove that (X, T) is a regular space. Let x  X be
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arbitrary and G a nbd of x.
By definition of locally compact space,
 a compact set A  X s.t. x  Aº  A  G.

A is compact, X is T -space  A is closed.
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