You know that f is continuous if and only if , right?
I would assume that and then show that . Or assume approaches a limit distinct from and then derive a contradiction.
Let M, N be metric spaces with metrics dm and dn respectively.
Let f be a function from M to N.
Suppose that if (an) is any convergent sequence in M, then (f(an)) is a convergent sequence in N.
Show that f is a continuous function.
Note that the question does not say that (an) --> a implies (f(an)) --> f(a). If it did then I could solve the question. But, as it is currently stated, I do not know the answer.
But, here's what I was working on before, and you can feel free to give me your thoughts.
Suppose f is not continuous:
Then, there exists e>0 and a xn (n=1,2,3,...) in M where
dM(xn, a) < 1/n but dN(f(xn), f(a)) > e
clearly (xn) converges to a, so by assumption (f(xn)) is a convergent sequence. Say it converges to f(x)
Thus, for n large enough, we have dN(f(xn), f(x))< e + d(f(x),f(a)) and dM(an, a)<1/n
but d<(xn, a) < 1/n --> e < dN(f(xn), f(a)) <= d(f(xn), f(x)) + d(f(x),f(a))
--> dN(f(xn), f(a)) > e - dN(f(x),f(a))
--> e + d(f(x),f(a)) > e + dN(f(x),f(a))
--> d(f(x),f(a)) > 0
--> f(x) /= f(a)
And I'm stuck
(well, not really, I could use the last posters hint to say f(x) = f(a), but in that case all this crap above would be unnecessary)
So, now suppose that but . Then, since any metric space is Hausdorff we may find open sets such that but , and since we have that contains all but finitely many points of . Now the definition for means that every neighborhood of there exists a neighborhood of such that . But, given any neighborhood of we have that there exists all but finitely many points of and so it follows that . This is clearly contradicts that . It follows that .
Note: It is really late here and I may have a) made a mistake b) missed a much easier method. Also, I assumed that the limit exists...and it is easy to prove that it must.