I know preimage of a dense set under open map is dense. Is it true that preimage of a dense set under continuous onto function is dense?

Actually when I did the problem that continuous onto image of a dense set is dense. Suddenly I thought that question. Is that true? If so, then how to prove?


Not true. Consider the identity map $id: \mathbb{R} \rightarrow \mathbb{R}$ where the first $\mathbb{R}$ is endowed with the discrete topology and the second, the euclidean topology. This map is continuous and surjective.

$\mathbb{Q}$ is dense in $\mathbb{R}$ under the Euclidean topology. Its inverse image under $id$ is also $\mathbb{Q}$. But $\mathbb{Q}$ is not dense in $\mathbb{R}$ under the discrete topology.

EDIT: OP asked the following: If we allow both $X$ and $Y$ to be $\mathbb{R}$ with the usual Euclidean topology. Will the hypothesis be true? The answer is no.

Consider the function: $f: \mathbb{R} \rightarrow \mathbb{R}_{\ge 0}$ by setting $f(x) =0$ if $x<0$ and $f(x)=0$ if $x\ge 0$. This map is continuous and surjective. The set $D=(0,\infty)$ is dense in $\mathbb{R}_{\ge 0}$. But its preimage under $f$, which is still $(0,\infty)$, is not dense in $\mathbb{R}$! A huge reason is because the map $f$ is not open.

  • $\begingroup$ Wow Great, thanks $\endgroup$ – Siraj Jun 12 at 15:26
  • $\begingroup$ is it true for real analysis i. e, if we take domain and codomain are $\mathbb{R}$ with usual topology then result is true? $\endgroup$ – Siraj Jun 12 at 15:41
  • $\begingroup$ Hi @Siraj I have edited my answer to include one more counter example. $\endgroup$ – thedilated Jun 12 at 16:02

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.