Metric and Topological Spaces

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## Continuity in metric spaces

We look at continuity for maps between metric spaces .

Definition
A map f between metric spaces is continuous at a point p X if
Given > 0  > 0 such that dX(p, x) <  dX(f(p), f(x)) < .

Informally: points close to p (in the metric dX) are mapped close to f(p) (in the metric dY).
A continuous function is one which is continuous for all p X.

Remarks
When one is given a point p and > 0 the one needs for the definition may depend on both p and . It is therefore incorrect to define continuity as: p X, > 0  > 0 such that x X with dX(p, x) <  dX(f(p), f(x)) < .
since this would imply that the same choice of would work for all p.

As in the prototype R case, one can connect continuity and convergence with:
Theorem
Continuous functions map convergent sequences to convergent sequences.

Formally if f: X Y is a map between metric spaces which is continuous and (an) is a sequence in X which is
convergent to a point p X them (f(an)) is a sequence in Y convergent to f(p).

Proof
Points close to p are mapped close to f(p).
More rigorously: to prove (f(an)) f(p), given > 0, use continuity at p X to find > 0 such that if dX(p, x) < then dY(f(p),f(x)) < .
Then use convergence in X to find N N such that if n > N we have dX(xn,p) < . For this N we have dY(f(xn), f(p)) < and so we have convergence in Y. In fact, as in the R case, there is a converse to this theorem.

Converse
If all convergent sequences are mapped to convergent sequences then the function is continuous.

More exactly: If (xn) x (f(xn)) f(p) then f is continuous at p.

Proof
Suppose that f were not continuous at p. Then for some > 0 we cannot find any choice of to satisfy the continuity condition.
In particular, = 1 will not work. Hence for some point x1 we have dX(x1 , p) < 1 but dY(f(x1), f(p))  .
Similarly = 1/2 will not work and so for some point x2 we have dX(x2 , p) < 1/2 but dY(f(x2), f(p))  , ...
Continue like this to get a sequence (x1 , x2 , x3 , ...) with dX(xn , p) < 1/n but dY(f(xn), f(p))  for each n. But since (xn) has been constructed so that (xn) p this contradicts the condition given in the theorem. Remark
Since "nice behaviour on convergent sequences" is a necessary and sufficient condition for continuity, this can be used as the definition of a continuous function.

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JOC February 2004