Map between two metric spaces and their limit

I have to proof the following:

Let $f: V \to W$ be a map between two metric spaces. Proof that $f(a)$ with $a \in V$ is continuous if and only if, for every converging sequence $x_n$ in V with limit $a$ the sequence $f(x_n)$ in W converges to $f(a)$.

First I tried to proof the $\Leftarrow$ statement. But I'm stuck.
Suppose we have the sequence $x_n$ with limit $a$ and $f(x_n)$ with limit $f(a)$. Let $\epsilon >0$, then $\exists N_1 \in \mathbb{N}$ such that if $n>N_1$ then $d(x_n,a)<\epsilon$ and $\exists N_2 \in \mathbb{N}$ s.t. if $n>N_2$ then $d(f(x_n),f(a))<\epsilon$. Take those $N_1, \ N_2$ such that this holds and take $N = \max\{N_1,N_2 \}$ then if $n>N$ the following is true $d(x_n,a)< \epsilon$ and $d(f(x_n),f(a))<\epsilon$.

I sort of proofed the existence of the limit with this proof but I still use $x_n$ which feels weird to me, I shouldn't have sequences in my final answer. Also for the other way around I have no clue on how to proceed as I have sort of the same problem with getting the sequence.

• Try to do it contradiction, assume $f(x_n) \not\to f(a)$ and then see why the $\epsilon\delta$-Criterion is not working. And the other way is essentially the same. – mjb4 May 25 '15 at 13:08

We have two metric spaces $(V, d)$ and $(W, \varrho)$. The proposition is the following:
$f: V \to W$ is continuous if and only if for each $a \in V$ and each sequence sequence $(x_n)_{n \in \Bbb N^\times}$ in $V$ with $\displaystyle \lim_{n \to \infty} x_n = a$ it holds that $\displaystyle \lim_{n \to \infty} f(x_n) = f(a)$.
Let's first assume that $f$ is continuous. Let $a \in V$ and $(x_n)_{n \in \Bbb N^\times}$ be a sequence in $V$ with $\displaystyle \lim_{n \to \infty} x_n = a$. We have to show that $\displaystyle \lim_{n \to \infty} f(x_n) = f(a)$. Let $\epsilon > 0$. Since $f$ is continuous in $a$, there exists a $\delta > 0$, such that for each $x \in V$ with $d(a,x) < \delta$ we have $\varrho(f(a), f(x)) < \epsilon$. Now since $\displaystyle \lim_{n \to \infty} x_n = a$, we find a $N \in \Bbb N^\times$, such that $d(x_n, a) < \delta$ for each $n \geq N$. So it follows that $\varrho(f(a), f(x_n)) < \epsilon$ for each $n \geq N$. Since $\epsilon > 0$ has been chosen arbitrary, we deduce that $$\lim_{n \to \infty} f(x_n) = f(a) \; .$$
Now assume that for $a \in V$ and each sequence $(x_n)_{n \in \Bbb N^\times}$ in $V$ with $\displaystyle \lim_{n \to \infty} x_n = a$ it holds that $\displaystyle \lim_{n \to \infty} f(x_n) = f(a)$. We show by a contradiction argument, that $f$ is continuous in $a$. So let's assume that $f$ is not continuous in $a$. Then by definition of continuity, there exists a $\epsilon > 0$, such that for each $\delta > 0$ we find a $x \in V$ with $d(a,x) < \delta$, but $\varrho (f(a), f(x) ) \geq \epsilon$. So for each $n \in \Bbb N^\times$ we find a $x_n \in V$ with $d(a, x_n) < \frac 1 n$, but $\varrho (f(a), f(x_n)) \geq \epsilon$. So this construction gives us a sequence $(x_n)_{n \in \Bbb N^\times}$ with $\displaystyle \lim_{n \to \infty} x_n = a$, but $\displaystyle \lim_{n \to \infty} f(x_n) \neq f(a)$, which is a constradition. So $f$ is continous in $a$.
• Start with $n = 1$. We find a $x \in V$ with $d(a,x) < 1$ and $\varrho(f(a), f(x)) \geq \epsilon$. Now set $x_1 := x$. Next, consider $n = 2$. We find a $x \in V$ with $d(a,x) < 1$ and $\varrho(f(a), f(x)) \geq \epsilon$. Set $x_2 := x$. By continuing this process, we have defined a sequence $(x_n)_{n \in \Bbb N^\times}$ with $\lim_{n \to \infty} x_n = a$ and $\lim_{n \to \infty} f(x_n) \neq f(a)$. – aexl May 29 '15 at 15:38