Let $f$ be a mapping of topological space $X$ to topological space $Y$. It's straightforward to check that if $f$ is continuous at $p$ we have $f(p_n) \to f(p)$ for all $p_n \to p$. Is the converse true? I know it's true for metric spaces: suppose $f$ is discontinuous at $p$ and construct $p_n \to p$ with $f(p_n) \not\to f(p)$ by choosing "bad" points from $1/n$-neighborhoods of $p$. However, this doesn't work if $X$ is not a metric space.
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This doesn't work in general. Consider the identity function $\text{id} : (\Bbb R, \mathcal T_{cc}) \to (\Bbb R, \mathcal T_{disc})$ where $\mathcal T_{cc}$ denotes the cocountable topology (the countable sets are exactly the closed sets [and $\Bbb R$ is also closed]), and $\mathcal T_{disc}\;$ denotes the discrete topology (every set is open).
Clearly, $\text{id} $ is not continuous. However, if $x_n \to x$ in $(\Bbb R, \mathcal T_{cc})$ then you can show that $x_n=x$ for $n$ large enough. Therefore, $\text{id}(x_n) = x_n \to x$ in $(\Bbb R, \mathcal T_{disc}\,)$.
See also here.
In general, the following holds : if $X$ is first countable and if $f:X \to Y$ is sequentially continuous (i.e. $x_n \to x$ in $X$ implies $f(x_n) \to f(x)$ in $Y$, for every convergent sequence $(x_n) \subset X$), then $f$ is continuous.
You can generalize this by considering nets.
