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?

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

1 Answer 1


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.


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