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This is the original question:

Let $f: X\to Y$ be a map between metric spaces. Prove that if $(f(x_n))_{n=0}^\infty$ converges in $Y$ whenever $(x_n)_{n=0}^\infty$ converges in $X$ then $f$ is continuous. [Note: it is not given that $f(x_n) → f(x)$ whenever $x_n → x$.]

I have spent quite a bit of time on this problem. I finally looked it up and have been reading the solution in the link below:

for every convergent sequence $x_n$, $f(x_n)$ also converges. Does this imply continuity of f?

I am confused by the solution given. I understand that they proved that $f(x_n)$ does not converge to $f(x)$, but this doesn't prove that $f(x_n)$ does not converge at all. If we want to use the contrapositive, shouldn't be have to show that if $f$ is discontinuous, then there is some sequence $(x_n)$ that converges in $X$ but $f(x_n)$ does not converge in $Y$. That is, we have to show that for some convergent sequence, there is some $\varepsilon$ such that for all $n$ greater than some $N$, $f(x_n)>\varepsilon$.

Any clarification or hints would be greatly appreciated.

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  • $\begingroup$ A convergent sequence in a metric space is a Cauchy sequence. But the sequence $f(x),f(x_1),f(x),f(x_2),...$ which was given in the accepted answer is clearly not Cauchy, because $d(f(x),f(x_n))\geq\epsilon$ for all $n$. So it doesn't converge at all. $\endgroup$
    – Mark
    Aug 9, 2019 at 21:12

1 Answer 1

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What I proved in that answer was that, if $f$ was discontinuous at some point $x$, then there is a sequence $(x_n)_{n\in\mathbb N}$ of elements of $X$ converging to $x$ such that the sequence $\bigl(f(x_n)\bigr)_{n\in\mathbb N}$ is not convergent. But we are assuming that no such sequence exists.

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    $\begingroup$ That is correct, yes, but that was not the approach that I had in mind. What I thought was that my sequence has a subsequence which converges to $f(x)$ and another subsequence such that each of its terms is at a distance at least equal to $\varepsilon$ (for a fixed number $\varepsilon>0$) from $f(x)$. This later subsequence cannot possibly converge to $f(x)$. But when a sequence converges, each of its subsequences converges to the same limit. $\endgroup$ Aug 9, 2019 at 21:24
  • $\begingroup$ That makes sense. Thank you for the help. $\endgroup$
    – smileemote
    Aug 9, 2019 at 21:29

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