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Let $X$ and $Y$ be topological spaces, and let $x_1,...,x_m$ be distinct points in $X$. Let $f: \{x_1,...,x_m\} \to Y$ be a function.

Is there a continuous extension of $f$ from $\{x_1,...,x_m\}$ to $X$?

It's easy to see that this is true if $X = Y = \mathbb R$, for example. But I'm not sure how far this generalizes. Does the result hold for arbitrary topological spaces, or do we need some assumptions about $X$ and $Y$?

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It does not hold for arbitrary topological spaces if $m>1$: there are spaces $X$ with the property that every real-valued continuous function on $X$ is constant. Thus, if $x$ and $y$ are distinct points of such a space $X$, $r$ and $s$ are distinct real numbers, $f'(x)=r$, and $f'(y)=s$, there is no continuous function $f:X\to\Bbb R$ extending $f'$. It’s not hard to construct such counterexamples if we don’t impose any ‘niceness’ conditions on $X$ — just give $X$ the indiscrete topology, for instance — but in this answer I describe a construction, due to Eric van Douwen, of such spaces starting with any $T_3$-space containing two points that cannot be separated by a continuous real-valued function; in this answer I describe such a $T_3$-space, due to John Thomas.

If $X$ is a $T_4$-space, the Tietze extension theorem ensures that every real-valued function on a finite subset of $X$ has a continuous extension to $X$.

The accepted answer to this question is also relevant.

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  • $\begingroup$ The Tietze extension theorem also needs $Y = \mathbb R$, right? $\endgroup$ – user435571 Jul 11 at 2:36
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    $\begingroup$ @user435571: Yes: that’s why I said real-valued. The last link in my answer gives more information about what can happen when $Y\ne\Bbb R$. $\endgroup$ – Brian M. Scott Jul 11 at 2:38
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This is false. Let $X$ and $Y$ have at least $2$ points. Take $X$ to have the indiscrete topology and suppose that $y \in Y$. Take some $f$ with $y \in im(f)$ and $|im(f)| \geq 2$. Then let $U = Y - \{y\}$. If there was a continuous extension $F$, $F^{-1}[U]$ must then by nonempty and open, hence it must be all of $X$. Thus, all of $F$ must map into $U$, but we assumed that $y \in im(f)$, a contradiction.

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