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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.

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  • $\begingroup$ 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. $\endgroup$ – mjb4 May 25 '15 at 13:08
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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 $a$.

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  • $\begingroup$ The proof is clear to me, but I wonder if the way you define your sequence in the second part of the proof is good enough. During an exercise class the teach assistant said that you have to really define the sequence. Obviously you have your sequence in V but where do you really define this sequence? $\endgroup$ – DeanTheMachine May 29 '15 at 11:37
  • $\begingroup$ 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 $\lim_{n \to \infty} f(x_n) \neq f(a)$. $\endgroup$ – aexl May 29 '15 at 15:38

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