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Let $X$ be a sequential space, $Y$ be some topological space, and $f:X\mapsto Y$ define some function. If $\forall x_n \to x, n\in\mathbb{N}$ implies $f(x_n) \to f(x)$, then does it follow that $f$ is continuous at the point $x$?

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1 Answer

No.

Consider the Arens´ space $X=\{\infty\} \cup \mathbb{N} \cup (\mathbb{N}\times \mathbb{N})$, where each $(n,m)$ is isolated, basic neighborhoods of $n$ are of the form $B_{n,k}:=\{(n,m):m \geq k \} \cup \{n\}$ for some $k \in \mathbb{N}$ and basic neighborhoods of $\infty$ are obtained by removing from $X$ finitely many sets of the form $B_{n,0}$ and finitely many points from each of the remaining $B_{n,0}$´s. Let $f:X \to \mathbb{R}$ be the function defined by: $f((n,m))=1$, $f(n)=\frac{1}{n+1}$ and $f(\infty)=0$. This function satisfies your condition at $x:= \infty$ but is not continuous at $x$.

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