I have to show that there is no injective continuous map from $(\mathbb{R}-\mathbb{Q})\times\mathbb{R}$ to $\mathbb{R}$. Let $Y=(\mathbb{R}-\mathbb{Q})\times\mathbb{R}$.

I thought about doing something with connectedness and the image of a connected set by a continuous function (supposing $f$ exists) and lead to a contradiction. However, I tried looking at preimages of open disjoint sets in $\mathbb{R}$ but that would give me also open sets in the domain. Maybe removing a point from codomain $\mathbb{R}$ and get a connected (component) in domain?

  • 4
    $\begingroup$ If you get uncountably many disjoint intervals in $\mathbb{R}$, it's a contradiction since you can label each interval with a rational number, and since $\mathbb{Q}$ is $\aleph_0$ it isn't possible to have an injectivity from $\aleph_1\to\aleph_0$? $\endgroup$
    – Fabrizio G
    Jun 2, 2022 at 19:03
  • $\begingroup$ @FabrizioGambelín You should now write an answer to your own question. $\endgroup$
    – Paul Frost
    Jun 3, 2022 at 22:27
  • 1
    $\begingroup$ Yes I will, I had no time to do it, soon it will appear. $\endgroup$
    – Fabrizio G
    Jun 3, 2022 at 22:43


You must log in to answer this question.

Browse other questions tagged .