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.
Alternatively to Opalg's (this is true in any topological space) but let be an arbitrary neighborhood of then is a neighborhood of (by 's continuity) and so we see that is an open set containing and since it contains all but finitely many of so then contains all but finitely many values of .
Well, I'll tell you off the bat I have know idea what "strike" represents
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)
How about this. I am going to assume a lot of stuff and let you generalize! Mainly that you call a function continuous at if .
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.