Metric and Topological Spaces

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Continuity in metric spaces

We look at continuity for maps between metric spaces .

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

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 belongs X.

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

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

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

Points close to p are mapped close to f(p).
More rigorously: to prove (f(an)) rarrow f(p), given epsilon > 0, use continuity at p belongs X to find delta > 0 such that if dX(p, x) < delta then dY(f(p),f(x)) < epsilon.
Then use convergence in X to find N belongs N such that if n > N we have dX(xn,p) < delta. For this N we have dY(f(xn), f(p)) < epsilon and so we have convergence in Y.

In fact, as in the R case, there is a converse to this theorem.

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

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

Suppose that f were not continuous at p. Then for some epsilon > 0 we cannot find any choice of delta to satisfy the continuity condition.
In particular, delta = 1 will not work. Hence for some point x1 we have dX(x1 , p) < 1 but dY(f(x1), f(p)) gte epsilon.
Similarly delta = 1/2 will not work and so for some point x2 we have dX(x2 , p) < 1/2 but dY(f(x2), f(p)) gte epsilon, ...
Continue like this to get a sequence (x1 , x2 , x3 , ...) with dX(xn , p) < 1/n but dY(f(xn), f(p)) gte epsilon for each n. But since (xn) has been constructed so that (xn) rarrow p this contradicts the condition given in the theorem.

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