0
$\begingroup$

I have to define open map from $\mathbb{R}$ to $\{0,1\}$ (Sierpinski). I need it so I can show that there is homeomorphism from $\mathbb{R}/\mathbb{R}\setminus\{0\}$. That part is very easy with theorems that we have.

My mapping takes x into $0$, if $x\in ]-\infty,0[$ or $x\in]0,\infty[$ and then $x$ to $1$, if $x=0$.

It is continuous and surjective.

I say that it's open, because if $U\subseteq \mathbb{R}$ is open, then $f(U)=\{0,1\}$, $f(U)=\emptyset$ or $f(U)=\{0\}$ which are open sets in $\{0,1\}$.

Am I correct?

$\endgroup$
1
  • $\begingroup$ yes you are correct $\endgroup$
    – Anguepa
    Nov 14, 2015 at 12:08

1 Answer 1

4
$\begingroup$

This is indeed correct. An open set of the reals cannot contain only $x=0$, as $\{x\}$ is not open. So the possible images of open sets are correct.

Generalise: if $X$ is $T_1$ and has at least one point $p$ that is not an isolated point, then there is an open map from $X$ onto the Sierpinski space, as we map $p$ to $1$ and $X \setminus \{p\}$ to $0$. Then $f$ is onto, open en continuous.

$\endgroup$

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.