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Please, don't write the entire answer. I am looking for hints only.

Here $\mathbb{R}_S\times\mathbb{R}_S$ is the Sorgenfrey plane.

My attempt so far was limited by

  • Suppose it is continuous and it has a non-empty interior. Let $p$ be an interior point. Define $g(x)=f(x)-p$ so that $(0,0)$ belongs to its interior now. The diagonal of the Sorgenfrey plane has some properties, so I was thinking about using it.

  • Suppose you could extend it to $\mathbb{R}$, so $\tilde{f}(\mathbb{R})$ would be connected. However, if $A$ is a basic open set inside the image, one can show that it is not connected and the question is finished.

Could you help me? :) Remember, just a hint.

Cheers.

B. Rivas
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2 Answers2

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HINT: Your idea of using the (reverse) diagonal is a good one. Any line segment $L$ in the plane with slope $-1$ is an uncountable closed, discrete set in the Sorgenfrey plane. What does that tell you about $f^{-1}[L]$ in the irrationals?

Brian M. Scott
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  • Hi, Brian. That is exactly my problem. I stopped at this point. What does it tell? Should I insist a little more to see clearer? I will think a little more about it. Thank you. – B. Rivas May 07 '16 at 16:01
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    @B.Rivas: The inverse image of a closed set is closed, and it’s not hard to check that the inverse image of a discrete set is discrete; does that help? – Brian M. Scott May 07 '16 at 16:03
  • Oh, I think I finally get it. I didn't know about the inverse image of discrete sets. In this way, I have a discrete set inside the irrationals, but you can't have it, right? Do I need the closed part? – B. Rivas May 10 '16 at 11:27
  • @B.Rivas: True, you don't actually need the fact that the inverse image would be closed, since any discrete subset of the irrationals is countable. This follows from the fact that the irrationals are second countable. – Brian M. Scott May 10 '16 at 12:57
  • @Briam M. Scott: Hi again. I'm having some trouble proving that inverse image of a discrete set is discrete. For instance, if $f: \mathbb{R} \to \mathbb{N}$ is constant, then its inverse image is not discrete. Help. – B. Rivas May 10 '16 at 15:50
  • @B.Rivas: For each $x\in L$ there is an open nbhd $V_x$ of $x$ such that $V_x\cap L={x}$. Pick any $y_x\in\Bbb R\setminus\Bbb Q$ such that $f(y_x)=x$, and let $D={y_x:x\in L}$. Then $f^{-1}[V_x]$ is an open nbhd of $y_x$ in $\Bbb R\setminus\Bbb Q$, and $D\cap f^{-1}[V_x]={y_x}$. – Brian M. Scott May 10 '16 at 15:54
  • So I have to eliminate the case where $f$ is constant, right? Otherwise we would get the empty set at the end. ($f$ constant the exercise is clear) – B. Rivas May 10 '16 at 16:02
  • @B.Rivas: The point is that every non-empty open subset of the Sorgenfrey plane contains a segment $L$ with slope $-1$, so if $f[\Bbb R\setminus\Bbb Q]$ has non-empty interior, it must map onto such a segment. You’re never dealing with the constant case. – Brian M. Scott May 10 '16 at 16:04
  • I disagree slightly with that argument because our function is not necessary injective. I'm writing an answer, do you mind reading it later? Cheers. – B. Rivas May 11 '16 at 23:19
  • @B.Rivas: Look at it more carefully: the argument does not require $f$ to be injective. – Brian M. Scott May 11 '16 at 23:21
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Denote the irrationals by $\mathbb{P}$. Suppose $f$ is continuous and it has a non-empty interior. Let $L$ denote the segment of anti-diagonal in the interior.

The anti-diagonal of the Sorgenfrey Plane is discrete and uncountable, hence $\mathscr{C}=(f^{-1}(x))_{x \in L}$ is an uncountable and pairwise disjoint open cover for $f^{-1}(L) \subset \mathbb{P}$. The existence of $\mathscr{C}$ is a contradiction due to the fact that $f^{-1}(L) \subset \mathbb{R}$ is second countable, hence there are no uncountable and pairwise disjoint open cover.

B. Rivas
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