1
$\begingroup$

Let $(X,T_1),\,(Y,T_2)$ be topological spaces and $H$ be a homeomorphism from $X \to Y$. If $C$ is a dense subset of $X$, is $H(C)$ necessarily dense in the $Y$?

I thank you in advance.

$\endgroup$
  • 1
    $\begingroup$ What if $X$ is a space with just one point? $\endgroup$ – Ethan Bolker Jul 9 '17 at 14:36
  • 3
    $\begingroup$ @EthanBolker Then $Y$ would also have to be a space with just one point, and $C$ would have to be all of $X$. $\endgroup$ – Daniel Schepler Jul 9 '17 at 14:42
  • $\begingroup$ @DanielSchepler My mistake. I was planning to give the first sentence of your answer (upvoted) but then misthought that a homoemorphism need not be surjective. $\endgroup$ – Ethan Bolker Jul 9 '17 at 16:17
3
$\begingroup$

Since density is a property which only depends on the topology, you should of course expect this to be true.

Namely, suppose $U$ is a nonempty open subset of $Y$. Then since $H$ is bijective, we can rewrite $U \cap H(C) = H(H^{-1}(U) \cap C)$. However, $H^{-1}(U)$ is open by continuity of $H$, and nonempty since $H$ is surjective. Therefore, $H^{-1}(U) \cap C$ is nonempty since $C$ is dense; and therefore, $U \cap H(C) = H(H^{-1}(U) \cap C)$ is also nonempty.

(So, this actually proves your proposition only using that $H$ is continuous and bijective. It is actually possible to refine the proof to work only assuming that $H$ is continuous and surjective - in that case, $U \cap H(C) \supseteq H(H^{-1}(U) \cap C)$.)

$\endgroup$
  • $\begingroup$ I thank you so much but being continuous isn't enough. $\endgroup$ – user460818 Jul 9 '17 at 15:06
  • $\begingroup$ @RollingStone Nobody said it is. WHat Daniel Schepler wrote was that continuity and being bijective is enough, and he is right. $\endgroup$ – José Carlos Santos Jul 9 '17 at 15:09
  • $\begingroup$ yes mistake was from me, I'm sorry. $\endgroup$ – user460818 Jul 9 '17 at 15:21
1
$\begingroup$

Yes. Let $y\in Y$; you want to prove that, for every neighborhood $N$ of $y$, $N\cap H(C)\neq\emptyset$. Take $x\in X$ such that $f(x)=y$. Then $f^{-1}(N)$ is a neighborood of $x$ and therefore $f^{-1}(N)\cap C\neq\emptyset$. So, $N\cap H(C)\neq\emptyset$.

$\endgroup$
  • $\begingroup$ Professor Jose Carlos Santos, I thank you so much. $\endgroup$ – user460818 Jul 9 '17 at 15:07

Your Answer

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