# Proving that a mapping between metric spaces is convergent if it has a convergent sequence and its image is convergent.

I'm having some doubts whether the following proof is correct.

Let $(V,d_1)$ en $(W,d_2)$ be metric spaces and let $f:V\rightarrow W$ be a mapping from $V$ to $W$. Also for every convergent sequence $(x_n)_{n\geq 1}$ in $V$ with limit $a$, the sequence $(f(x_n))_{n\geq > 1}$ is convergent in $W$ with limit $f(a)$.

I want to prove that $f$ is continuous in $x=a$.

We know $\forall\epsilon^{'}>0\ \exists N_1\geq\mathbb{N}\ \forall n\geq N_1$ such that $(x_n)\in B(a,\epsilon^{'})$.
Also, $\forall\epsilon^{'}>0\ \exists N_2\geq\mathbb{N}\ \forall n\geq N_2$ such that $f(x_n)\in B(f(a),\epsilon^{'})$.
Combining gives us $\forall\epsilon^{'}\ \exists N\geq\mathbb{N}\ \forall n\geq N$ such that $f(B(a,\epsilon^{'}))\subset B(f(a),\epsilon^{'})$

Now let $\epsilon>0$ be given. Choose $\delta=\epsilon^{'}=\epsilon$. Now, for all $x\in B(a,\delta)=B(a,\epsilon^{'})$ we know that $f(x)\in B(f(a),\epsilon^{'})=B(f(a),\epsilon)$. So $f(B(a,\delta))\subset B(f(a),\epsilon)$. So $f$ is continuous.

I don't know if I can state this, because both sequences depend on an $N$. Does $x$ need to be an element in $(x_n)$, because then it does not hold for all $\in V$, right?

• I always like to consider contrapositives. In this case, that means to assume $f$ is not continuous, and use that to construct a sequence $x_n\to a$ such that $f(x_n)$ doesn't converge to $f(a)$. I haven't checked all the details, but it seems like it ought to be easier. – Arthur Jun 21 '18 at 22:28

In this case, I don't think you can avoid the contrapositive. Because given any $\varepsilon>0$, at best you will obtain a $\delta$ for a fixed sequence, but you need it to work for any element in $B(a,\delta)$.
If, on the other hand, you assume that $f$ is not continuous at $a$, then there exists $\varepsilon>0$ such that for any $\delta>0$ there exists $x_\delta$ with $d(x_\delta,a)<\delta$ and $d(f(x_\delta),f(a))>\varepsilon$. Taking $\delta=1/n$ for each $n$, we found $x_n\in V$ with $d(x_n,a)<1/n$ and $d(f(x_n),f(a))>\varepsilon$. That way we get a sequence $\{x_n\}\subset V$ with $x_n\to a$ and $f(x_n)\not\to f(a)$.