1
$\begingroup$

Let $X$ be a Hausdorff space, $D \subset X$ be a dense set, and $f:X \rightarrow Y$ be a continuous function such that $f|_D:D \rightarrow f(D)$ is a homeomorphism.

Show that $f(X \setminus D) \subset Y \setminus f(D)$.


Here is what I've managed to show so far:

1.

$f(D)$ is dense in $f(X)$: Let $V \subset f(X)$ be an open set. Then $f^{-1}(V)$ is also open by continuity of $f$. $D$ is dense in $X$. Hence, there exists $d \in D \cap f^{-1}(V)$. So $f(d) \in f(D) \cap V$.

2.

$f(D)$ has the Hausdorff (H) property: $D \subset X$ has H in it's topology. By Homeomorphism, $f(D)$ has H as well.

$\endgroup$
  • $\begingroup$ I haven't thought it through, so this might be the totally wrong road to take: consider a proof by contradiction so that $E=f(X\setminus D)\cap f(D)\neq\varnothing$. Take $x\in E$, then by definition, there exists $y_1\in X\setminus D$ and $y_2\in D$ such that $f(y_1)=f(y_2)=x$. Now using the Hausdorff property, there exist $y_1\in U_1$ and $y_2\in U_2$ such that $U_1\cap U_2=\varnothing$ and $U_1,U_2$ are open. $\endgroup$ – Clayton Feb 13 '18 at 19:00
  • $\begingroup$ Why does the Hausdorff property hold on $Y$? @user284331 $\endgroup$ – mibarg Feb 14 '18 at 7:28
  • $\begingroup$ There is on $X$, $y_{1},y_{2}$ are elements in $X$. $\endgroup$ – user284331 Feb 14 '18 at 7:39
  • $\begingroup$ Sorry, I've misread your comment. Why does this lead to a contradiction? @user284331 $\endgroup$ – mibarg Feb 14 '18 at 9:00
1
$\begingroup$

If $f(x)=f(y)$ for $x\in X-D$, $y\in D$, then $f(x_{\delta})\rightarrow f(x)$ for a net $(x_{\delta})\subseteq D$ such that $x_{\delta}\rightarrow x$, so $f(x_{\delta})\rightarrow f(y)$. Since $f|_{D}$ is a homeomorphism, then $x_{\delta}\rightarrow y$, but $X$ is Hausdorff, so $x=y$, a contradiction.

$\endgroup$
  • $\begingroup$ I'm unfamiliar with nets. Is this question solvable without them? $\endgroup$ – mibarg Feb 14 '18 at 7:20
  • $\begingroup$ I believe what @Clayton wrote there is what you are looking for? $\endgroup$ – user284331 Feb 14 '18 at 7:21

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.