Let $X$, $Y$ be topological spaces and let $f\colon E\to Y$ be continuous where $E\subseteq X$. Let $S\subseteq \overline E\setminus E$ and $g\colon S\to Y$ be such that $f(x)\to g(c)$ whenever $x\to c$ (according to the first definition here). Now, extend $f$ to $h\colon E\cup S\to Y$ via $g$.

Question: Is $h$ continuous?

The answer is shown to hold when $X$, $Y$ are taken to be metric spaces, $f$ taken to be uniformly continuous, and $S = \overline E\setminus E$.

  • $\begingroup$ There are two definitions in the section your first link links to. It's not entirely clear which one you mean – "whenever" sounds to me a bit like you might mean the sequential limit. (The two definitions are equivalent in metric spaces, so we can't infer from the answer your second link links to which one you mean.) $\endgroup$
    – joriki
    Jan 16, 2023 at 15:31
  • $\begingroup$ @joriki I meant the first definition. $\endgroup$
    – Atom
    Jan 16, 2023 at 15:51
  • $\begingroup$ Is it even necessarily true that $g$ is continuous? $\endgroup$
    – FShrike
    Jan 16, 2023 at 17:43
  • $\begingroup$ @FShrike In $\mathbb R$, for $E$ being a (nonempty) open interval, $g$ is trivially continuous (domain being a two-point set). I can't comment on general case. $\endgroup$
    – Atom
    Jan 16, 2023 at 19:04

1 Answer 1


Under the assumption of regularity of $Y$ this is true. The proof below was derived from Bourbaki "General topology" (Theorem I.8.5.1). They also provide a counterexample for arbitrary non-regular space $Y$ (so given $Y$ they construct $X,E,f,g$ s.t. all your conditions are satisfied, but continuity of $h$ fails) - Exercise 19 in I.8. This counterexample is somewhat complicated (not mentioning that the construction of a non-regular Hausdorff space is already a nontrivial problem). Before reading the proof note that I use the term "neighborhood of a point" in the sense of Bourbaki, i.e. it means a set that contains this point in its interior.

Assume that $Y$ is regular (that is, for each $y \in Y$ and a neighborhood $U \subset Y$ of $y$ there is a closed neighborhood $F \subset Y$ of $y$ s.t. $F \subset U$) and consider $x \in E \cup S$. Let $F \subset Y$ be a closed neighborhood of $h(x)$. I claim that there exists an open set $U \subset X$ such that $x \in U$ and $f(U \cap E) \subset F$. Indeed, if $x \in E$, then it is true by continuity of $f$ and, if $x \in S$, then it is true by the assumptions on $f$ and $g$. It remains to show that $g(U \cap S) \subset F$. Let $z \in U \cap S$. Since $z \in \overline{E}$, it follows that in all neighborhoods of $z$ there are points from $U \cap E$, and by the assumption that $f(x) \rightarrow g(z)$, when $x \rightarrow z$, we obtain that $g(z) \in \overline{f(U \cap E)}$. So, $g(z) \in \overline{f(U \cap E)} \subset \overline{F} = F$. Therefore, $h(U \cap (E \cup S)) \subset F$ and by regularity of $Y$ (note that we considered only closed neighborhoods of $h(x)$) we obtain continuity of $h$.

  • $\begingroup$ Sorry for such a late reply. This was a great proof! $\endgroup$
    – Atom
    Jan 21, 2023 at 19:01

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