4
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
  • 3
    $\begingroup$ Then your book is completely wrong; throw it away. $\endgroup$ – Chris Eagle Jun 26 '13 at 16:06
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.