Continuous functions in metric spaces

**southprkfan1** · February 26th 2010, 11:11 AM

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.

**JG89** · February 26th 2010, 11:46 AM

You know that f is continuous if and only if $a_n \rightarrow a \Rightarrow f(a_n) \rightarrow f(a)$ , right?

I would assume that $f(a_n) \rightarrow c$ and then show that $c = a$ . Or assume $f(a_n)$ approaches a limit distinct from $a$ and then derive a contradiction.

**xalk** · February 26th 2010, 12:21 PM

what happened to plato post??

**Opalg** · February 26th 2010, 12:25 PM

Hint: Suppose that $a_n\to a$ and that $f(a_n)\to b$ . What can you say about the limit of the sequence $a_1,a,a_2,a,a_3,a,\ldots$ ?

**Drexel28** · February 26th 2010, 01:56 PM

Originally Posted by southprkfan1

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 $N$ be an arbitrary neighborhood of $f(a)$ then $f^{-1}(N)$ is a neighborhood of $a$ (by $f$ 's continuity) and so we see that $f^{-1}(N)$ is an open set containing $a$ and since $a_n\to a$ it contains all but finitely many of $\{a_n\}_{n\in\mathbb{N}}$ so then $ff^{-1}(N)\subseteq N$ contains all but finitely many values of $f(a_n)$ .

**Plato** · February 26th 2010, 02:09 PM

Originally Posted by Drexel28

$f^{-1}(N)$ is a neighborhood of $a$ (by $f$ 's continuity) .

But the point is to show that $f$ is continuous.
So we cannot assume that it is, can you?

**Drexel28** · February 26th 2010, 02:12 PM

Originally Posted by Plato

But the point is to show that $f$ is continuous.
So we cannot assume that it is, can you?

Oops! Misread the question. And well...I probably shouldn't give the full solution...but I do want to mention that this is true whenever the domain is a first countable space.

**southprkfan1** · February 26th 2010, 02:56 PM

Originally Posted by Opalg

Hint: Suppose that $a_n\to a$ and that $f(a_n)\to b$ . What can you say about the limit of the sequence $a_1,a,a_2,a,a_3,a,\ldots$ ?

Thanks for hint, I see how it implies that (f(an)) converges to f(a).

Just out of curiosity, is this really the best way to go about it? I would think there's a more "direct" route to take. Maybe I'm wrong.

Either way, thanks a lot!

**Drexel28** · February 26th 2010, 03:04 PM

Originally Posted by southprkfan1

Thanks for hint, I see how it implies that (f(an)) converges to f(a).

Just out of curiosity, is this really the best way to go about it? I would think there's a more "direct" route to take. Maybe I'm wrong.

Either way, thanks a lot!

There is, a way dealing with open balls around $f(x)$ and $x$ . Try writign out what continuity means and see if something doesn't "strike" you

**Plato** · February 26th 2010, 03:19 PM

Originally Posted by southprkfan1

I see how it implies that (f(an)) converges to f(a). Just out of curiosity, is this really the best way to go about it? I would think there's a more "direct" route to take.

That is the very defintion of continuity.
How can that be inproved upon?

**southprkfan1** · February 26th 2010, 03:27 PM

Originally Posted by Drexel28

There is, a way dealing with open balls around $f(x)$ and $x$ . Try writign out what continuity means and see if something doesn't "strike" you

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)

**Drexel28** · February 26th 2010, 10:29 PM

Originally Posted by southprkfan1

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 $f$ continuous at $x$ if $\lim_{\xi\to x}f(\xi)=f(x)$ .

So, now suppose that $x_n\to x\implies f(x_n)\to f(x)$ but $\lim_{\xi\to x}f(\xi)=y\ne f(x)$ . Then, since any metric space is Hausdorff we may find open sets $O_y,O_{f(x)}$ such that $y\in O_y,f(x)\in O_{f(x)}$ but $O_y\cap O_{f(x)}=\varnothing$ , and since $f(x_n)\to f(x)$ we have that $O_{f(x)}$ contains all but finitely many points of $f(x_n)$ . Now the definition for $\lim_{\xi\to x}f(x)=y$ means that every neighborhood $E$ of $y$ there exists a neighborhood $N$ of $x$ such that $f\left(N-\{x\}\right)\subseteq E$ . But, given any neighborhood $N$ of $x$ we have that there exists all but finitely many points of $\{x_n\}$ and so it follows that $f\left(E-\{x\}\right)\cap O_{f(x)}\ne\varnothing\implies f\left(E-\{x\}\right)\nsubseteq O_y$ . This is clearly contradicts that $\lim_{\xi\to x}f(\xi)=y$ . It follows that $\lim_{\xi\to x}f(\xi)=f(x)$ .

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.

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