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Could you give the counterexample for the following case:

Let $(X,\tau)$ be a topological space which is not first countable, then it is possible that $f$ is not continuous at some $x \in X$ but $f(x_n)$ converges to $f(x)$ for every sequence $(x_n)$ that converges to $x$

Thanks a lot!

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up vote 7 down vote accepted

Let $X$ be an uncountable set, and let $\tau$ be the co-countable topology on $X$. Let $\tau_d$ be the discrete topology on $X$. I’ll leave it to you to verify that the identity map on $X$, considered as a function from $\langle X,\tau\rangle$ to $\langle X,\tau_x\rangle$, has the desired properties. HINT: The only convergent sequences in either topology are the ones that are eventually constant.

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