Theorem
A continuous image of a compact set is compact.
Proof
Suppose
iscontinuous and
and
aretopological spaces, and
isa compact subset of![]()
Let
bean open cover of
sothat![]()
Then![]()
Since
iscontinuous, the sets
areopen and
isan open cover of![]()
iscompact, hence
isreducible to a finite subcover, say![]()
Then![]()
Hence![]()
Hence f(A) is compact.