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Let X a Hausdorff Tychonoff space, C a closed subset of X, A a compact subset of int C and B a closed subset of C such that A and B are disjoint. Let f: A---> [0,1] and g: B---> [0,1] be continuous functions. Is it true that there exist a continuous extension h with domain C of both functions f and g? We know that f has such an extension, by compacity of A and X being Tychonoff, can we do that extension in a way to be also an extension of g? Thank you in advance.

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    $\begingroup$ $X$ is totally irrelevant to the question, you might as well assume that $X=C$. $\endgroup$ – Eric Wofsey Nov 26 '19 at 5:42
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No. Indeed, such an extension need not exist even if $A$ is empty, so you are just asking for an extension of $g$. For instance, let $X=C$ be any Tychonoff space that is not normal (say, the deleted Tychonoff plank) and let $S$ and $T$ be disjoint closed subsets of $X$ that cannot be separated by open sets. Then the function $g$ on $B=S\cup T$ that is $0$ on $S$ and $1$ on $T$ cannot be extended continuously to $C$.

(Of course, if $X$ is normal then such an extension does always exist by the Tietze extension theorem.)

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  • $\begingroup$ Simpler examples are the best, thank you. What about a little modification, say, C being a proper subset of X, A the same and B the boundary of C $\endgroup$ – user78594 Nov 26 '19 at 6:00
  • $\begingroup$ You can just embed my example in a larger space such that $B$ becomes its boundary (for instance take $X=C\times\{0\}\cup B\times[0,1]$). $\endgroup$ – Eric Wofsey Nov 26 '19 at 6:05
  • $\begingroup$ Oh, I see. Better made up my mind before any minor modification of my original question. Thanks a lot. $\endgroup$ – user78594 Nov 26 '19 at 7:33

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