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Let $T,Y$ topological spaces with $Y$ a regular space, $X \subset T$ a dense subspace and $f: T \to Y$ a function with the property that $f\vert_{X \cup \{t \}}$ is continuous for all $t \in T$. Prove that $f$ is continuous.

My attempt. Let $t \in T$, then we have that $f\vert_{X \cup \{t \}}$ is continuous. Now let be $M$ an open neighborhood of $f(t)$, then $f(t)\notin Y\setminus M$, there exist open sets $U_1$ and $U_2$, such that $f(t) \in U_1$, $Y\setminus M \subset U_2$ and $U_1 \cap U_2 =\varnothing$. However, I am still not very clear how to use the fact that $X$ is dense.

Anne Bauval
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Alejandra Benítez
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    The answer can be found here (your assumption that $f|_{X \cup {t}}$ is continuous for any $t \in T$ implies the requirement therein): https://math.stackexchange.com/questions/459232/nets-dense-subsets-and-continuous-maps – Florian R Sep 04 '24 at 03:39

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