2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Arthur Jun 21 '18 at 22:28
2
$\begingroup$

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 $f(x_n)\not\to f(a)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.