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Let $X$ be a metric space and $Y$ a topological space. Then for any map $f: X\to Y$, $f$ is continuous at point $x\in X$ if for every sequence $\{x_n\}\subset X$, $\lim_{n\to \infty}x_n=x\Longrightarrow \lim_{n\to \infty}f(x_n)=f(x)$. My question is, if we weaken the condition and let $X$ be a common topological space, does the extended proposition still hold?

Thanks in advance.

Martin Sleziak
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Jihai Zhu
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    For $X,Y$ top. spaces, one can define $f: X \to Y$ to be sequentially continouos if for any convergent sequence $(x_n){n \in \mathbb{N}} \in X$ with $x{\infty} = \lim_{n} x_n \in X$, the sequence $(f(x_n)){n \in \mathbb{N}}$ converges to $f(x{\infty})$ in $Y$. One can show that in a general topological space continuity (in the sense of preimage of open sets is open) implies sequential continuity. The converse is false, in general. For counterexamples, see: https://math.stackexchange.com/questions/1419938/example-of-topological-spaces-where-sequential-continuity-does-not-imply-continu – ferhenk Oct 29 '20 at 16:42

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