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Let $Q,Q^c$ denotes be the set of all rational and irrational numbers in $\mathbb {R}$. $Y=\mathbb {R^2}\setminus(Q×Q^C )$ with the usual subspace topology of $\mathbb {R^2}$.find Number of connected components of $Y$?

My answer : i thinks $Y = \mathbb{R} \times \mathbb{R} \cup \mathbb{R} \times \mathbb{R}$. Since $Y$ is covered by lines in both coordinate directions, it is path connected. so $Y$ has one connected component.

is its correct/incorrect ??

Pliz tell me

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    $\begingroup$ Subspace topology? Do you mean $\setminus$ instead of $/$? And wouldn't $\Bbb R\times \Bbb R\cup \Bbb R\times \Bbb R$ simply be $\Bbb R\times \Bbb R$ again, with nothing removed? $\endgroup$ – Hagen von Eitzen May 15 '18 at 6:07
  • $\begingroup$ i mean \ @HagenvonEitzen,,sorry sir wites mistakes...i edit now $\endgroup$ – jasmine May 15 '18 at 6:10
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    $\begingroup$ I think you mean $\Bbb R\times \Bbb Q\cup \Bbb Q^c\times \Bbb R$. Then, yes, this shows that $Y$ is path-connected. $\endgroup$ – Hagen von Eitzen May 15 '18 at 6:21
  • $\begingroup$ that mean how many connected components are there in Y ???@HagenvonEitzen sir $\endgroup$ – jasmine May 15 '18 at 6:45
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Notice that

$Y =\mathbb{R}^2\backslash(\mathbb{Q}\times\mathbb{Q}^c) = (\mathbb{R}\times (\mathbb{R}\backslash \mathbb{Q}^c))\cup((\mathbb{R}\backslash \mathbb{Q})\times \mathbb{Q}^c) = (\mathbb{R}\times\mathbb{Q})\cup(\mathbb{Q}^c\times\mathbb{Q}^c).$

So, if $y=(a,b)\in Y$ and $b\in \mathbb{Q}^c$, then $a\in\mathbb{Q}^c$ and $\{y\}$ is the connected component of $y$.

If, $y'= (a,b)\in Y$ and $b\in\mathbb{Q}$, then $a$ can be any real number, so $\{(t,b),\ t\in \mathbb{R}\}$ is the connected componet of $y'$.

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    $\begingroup$ I think this is incorrect, since the space is path connected $\endgroup$ – Exodd Sep 26 '18 at 13:30

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