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Prove or disprove: $\lbrace (x,y)\in \mathbb{R}^{2}: x^2+y^3\in \mathbb{R}\setminus \mathbb{Q}\rbrace$ is disconnected with the usual topology in $\mathbb{R}^{2}$.

Anupam
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    Hint:$f:\mathbb{R}^2\to\mathbb{R}$ $f(x,y)=x^2+y^3$ is continous function – Myshkin Sep 01 '13 at 15:11
  • @ Senore: I also tried that but failed. Please provide some more hint. – Anupam Sep 01 '13 at 15:17
  • do you know Image of a connected set under a continous map must be connected? $\mathbb{R}^2$ is connected – Myshkin Sep 01 '13 at 15:21
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    Yeah, I know that. But we are not finding image of $\mathbb{R}^{2}$, isn't it? – Anupam Sep 01 '13 at 15:23
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    @SucharitaDeb what can you tell about the sets $x^2+y^3<2$ and $x^2+y^3>2$ – clark Sep 01 '13 at 15:23
  • @SucharitaDeb: Combine the comments from Clark & Seonore to get a contradiction. Or note that the inverse image of a disconnected set under a continuous function must itself be disconnected. – copper.hat Sep 01 '13 at 15:25

1 Answers1

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This is really just fleshing out the details in the comments.

Let $A=\{ (x,y)\in \mathbb{R}^{2}: x^2+y^3\in \mathbb{R}\setminus \mathbb{Q}\}$ and note that $f\colon A\rightarrow \mathbb{R}$ given by $f=g|_A$ where $g\colon\mathbb{R}^2 \rightarrow \mathbb{R} \colon (x,y)\mapsto x^2+y^3$ is continuous as it is the restriction of a continuous function to a subset with the subspace topology. The image of $f$ is the set of irrational numbers which is a disconnected set and so $A$ can not be connected as the image of a connected space is a connected subset of the codomain.

Dan Rust
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