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Let $X, Y$ be topological spaces, $D \subseteq X$, and $f: D \to Y$ a continuous function. Suppose that $a \in X \setminus D$ is a limit point of $D$, and there is some $y \in Y$ such that $f(x) \to y$ as $x \to a$. Is it always the case that $\bar{f}: D \cup \{a\}\to Y$, $x \mapsto \begin{cases} f(x) & \text{if } x \in D \\ y & \text{if } x = a \end{cases}$ is continuous? Amann's analysis claims that it is always continuous, but I have trouble proving it if $X$ is not Hausdorff. For instance if $x \in D$, isn't it possible that any neighborhood $U$ of $x$ (in $D \cup \{a\}$)will contain $a$, so that $f(U)$ might not be in the desired neighborhood of $f(x). $

juekai
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    Does not he only work with metric spaces? – Moishe Kohan Jun 23 '24 at 15:36
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    You do not need $X$ to be Hausdorff. But you need $X$ to be first countable if you want to use sequences. You can use nets, though. For nets, what you need for $f$ to be continuous at $a$ is that $x \rightarrow a \Rightarrow f(x) \rightarrow f(a)$. – André Caldas Jun 23 '24 at 19:01

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First of all, Amann in his book

Amann, Herbert; Escher, Joachim, Analysis I, Grundstudium Mathematik. Basel: Birkhäuser. xv, 445 S. (1998). ZBL0909.26001.

works only with metric spaces (which makes sense for an analysis textbook), hence, his claim about continuity of the extension is totally justified. For general topological spaces, the continuity claim (as you suspected) is simply false and it has nothing to do with the 1st countability. Here is an example.

Let $A$ be a nonempty set, $X:=A\sqcup \{b\}$. I will equip $X$ with trivial topology. Note that $b$ is a limit point of $A$ in $X$ (this has nothing to do with sequences, it simply means that every neighborhood of $b$ has nonempty intersection with $A$, which is, of course, true). Let $Y=\{y, z\}$ be the topological space whose open subsets are $\emptyset, \{z\}$ and $Y$. Let $f: A\to Y$ be the constant function, $f(A)=\{z\}$. Then $f(x)\to y$ as $x\to b$. Indeed, the only neighborhood of $y$ is $Y$ and $f^{-1}(Y)=A$, which is obviously a punctured neighborhood of $b$. Define the function $g: X\to Y$ by extending $f$ to $b$ via $g(b)=y$. I claim that $g$ is not continuous. Indeed, for the open subset $V=\{z\}$, $f^{-1}(V)=A$. But $A$ is not open in $X$.

Moishe Kohan
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