There is no injective continuous map from $(\mathbb{R}-\mathbb{Q})\times\mathbb{R}$ to $\mathbb{R}$.

Asked Jun 02 '22 at 18:53

Active Jun 02 '22 at 18:53

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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?

asked Jun 02 '22 at 18:53

Fabrizio Gambelín

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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 $\aleph_1\to\aleph_0$? – Fabrizio Gambelín Jun 02 '22 at 19:03
@FabrizioGambelín You should now write an answer to your own question. – Paul Frost Jun 03 '22 at 22:27
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Yes I will, I had no time to do it, soon it will appear. – Fabrizio Gambelín Jun 03 '22 at 22:43

There is no injective continuous map from $(\mathbb{R}-\mathbb{Q})\times\mathbb{R}$ to $\mathbb{R}$.

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