Urysohn's lemma says that if $X$ is a normal space, then for every two disjoint closed sets $F_{1},F_{2}\in X$, there exists a continuous function $f:X\to [a,b]\in\Bbb{R}$ such that $f(F_{1})=\{a\}$ and $f(F_{2})=\{b\}$.

Tietze Extension theorem says that for every such $f$, there exists a continuous function $f^*:X\to [a,b]$ such that $f^*|F_{1}$ and $f^*| F_{2}=f$.

I don't understand the difference! Why can't $f^*=f$? And if we assume $f^*\neq f$, are we just saying that there are two such continuous functions of the type mentioned in Urysohn's lemma?

Thanks in advance!

  • $\begingroup$ I don't know where you're getting these statements from, but "every such $f$" is just wrong. $\endgroup$ – Chris Eagle Jun 26 '13 at 16:02
  • $\begingroup$ "every such $f$" as in every such $f$ as constructed in Urysohn's lemma. This has been quoted verbatim. $\endgroup$ – fierydemon Jun 26 '13 at 16:05
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    $\begingroup$ Then your book is completely wrong; throw it away. $\endgroup$ – Chris Eagle Jun 26 '13 at 16:06

One can indeed obtain the result of UL via TET by setting $f|_{F_1} = a$ and $f|_{F_2} = b$. Also, in this Wikipedia article it is explicitly written that TET generalizes UL.

  • $\begingroup$ My book gives two proofs. One for Urysohn's lemma, and the other for Tietze Extension theorem. I feel one proof should suffice for both, as both of them pretty much state the same thing! Is this understanding wrong? $\endgroup$ – fierydemon Jun 26 '13 at 15:54
  • $\begingroup$ @AyushKhaitan: unfortunately, I don't know which book are you talking about. Here the proof of TET is obtained by using UL. $\endgroup$ – Ilya Jun 26 '13 at 15:56
  • $\begingroup$ @Ilya- what I meant was shouldn't the proof for UL directly imply TET, rather than being a component of the machinery used to prove TET? And I am reading "General Topoogy" by Pervin. $\endgroup$ – fierydemon Jun 26 '13 at 16:01
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    $\begingroup$ @AyushKhaitan: Urysohn's lemma uses two non-empty disjoint closed sets. What you wrote in your question is just an application of Tietze's extension theorem to two closed sets. But the theorem itself says that each continuous $f$ from a closed set $C$ to $\mathbb R$ can be extended to a continuous $\hat f$ from the entire space $X$ to $\mathbb R$. $\endgroup$ – Stefan Hamcke Jun 26 '13 at 16:03
  • $\begingroup$ @StefanH.- thanks! I think that spells it out perfectly. $\endgroup$ – fierydemon Jun 26 '13 at 16:13

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