John Stillwell - Naive Lie Theory.pdf - Index of

172 8 Topology
true that f −1 (“open”)=“open” when f is continuous, and so the argument
goes through.
A convenient property of continuous functions on compact sets is uniform
continuity. As always, a continuous f : S → T has the property
that for each ε > 0 there is a δ > 0 such that f maps a δ-neighborhood of
each point P ∈ S into an ε-neighborhood of f (P) ∈ T . We say that f is
uniformly continuous if δ depends only on ε, not on P.
Uniform continuity. If K is a compact subset of R m and f : K → R n is
continuous, then f is uniformly continuous.
Proof. Since f is continuous, for any ε > 0andanyP ∈ K there is a neighborhood
N δ(P) (P) mapped by f into N ε/2 ( f (P)). To create some room to
move later, we cover K with the half-sized neighborhoods N δ(P)/2 (P),
then apply compactness to conclude that K is contained in some finite
union of them, say
K ⊆ N δ(P1 )/2(P 1 ) ∪ N δ(P2 )/2(P 2 ) ∪···∪N δ(Pk )/2(P k ).
If we let
δ = min{δ(P 1 )/2,δ(P 2 )/2,...,δ(P k )/2},
then each point in K lies in a set N δ(Pi )/2(P i ) and each of the sets N δ(Pi )(P i )
has radius at least 2δ. I claim that |Q − R| < δ implies | f (Q) − f (R)| < ε
for any Q,R ∈ K ,so f is uniformly continuous on K .
To see why, take any Q,R ∈ K such that |Q − R| < δ and a half-sized
neighborhood N δ(Pi )/2(P i ) that includes Q. Then
|P i − Q| < δ and |Q − R| < δ,
so it follows by the triangle inequality that
|P i − R| < 2δ, and hence R ∈ N δ(Pi )(P i ).
Also, it follows from the definition of N δ(Pi )(P i ) that | f (P i ) − f (Q)| < ε/2
and | f (P i ) − f (R)| < ε/2, so
| f (Q) − f (R)| < ε,
again by the triangle inequality.
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