# A question about extentions of continuous functions on Tychonoff spaces

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.

• $X$ is totally irrelevant to the question, you might as well assume that $X=C$. – Eric Wofsey Nov 26 '19 at 5:42

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.)
• You can just embed my example in a larger space such that $B$ becomes its boundary (for instance take $X=C\times\{0\}\cup B\times[0,1]$). – Eric Wofsey Nov 26 '19 at 6:05