The exercise says: Let $X$ be a subspace of $\mathbb{R}^2$ given by points with at least one irrational coordinate. Show that $X$ is path connected.
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2$(\mathbb R\setminus\mathbb Q)\times\mathbb Q$ is not path connected, but $\mathbb R^2\setminus (\mathbb Q^2)$ is path connected. – Cheerful Parsnip Jun 04 '19 at 13:41
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Sorry, I committed a mistake. The exercise says: Let X be a subspace of $\mathbb{R}^2$ given by points with at least one irrational coordinate. Show that X is path connected. – Giusy Jun 04 '19 at 13:51
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1Hmm. Can $A\times B$ be connected if $B$ is not connected? – MPW Jun 04 '19 at 13:52
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Yes it can be possible – Giusy Jun 04 '19 at 13:56
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1Maybe you can edit your question then. – InfiniteLooper Jun 04 '19 at 13:58
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Can you write your space $X$ in a a form like $\mathbb R^2 \backslash ???$ – InfiniteLooper Jun 04 '19 at 13:59
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Please clarify your question. The set you have described is usually interpreted as the first set in @CheerfulParsnip's comment and it not connected. – copper.hat Jun 04 '19 at 14:02
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@Giusy, if your problem was about $X$, the space of points with at least one irrational coordinate, then $X=\mathbb{R}^2\setminus \mathbb{Q}^2$. You should edit that in your question. – Julian Mejia Jun 04 '19 at 14:09
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Hint: Let $\pi(x,y)=y$. How connected is $\pi\bigl((\mathbb R\setminus\mathbb Q)\times\mathbb Q\bigr)$?
José Carlos Santos
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Sorry, I committed a mistake. The exercise says: Let X be a subspace of $\mathbb{R}^2$ given by points with at least one irrational coordinate. Show that X is path connected. – Giusy Jun 04 '19 at 13:51
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Of all lines through $a\in X$, only countably hit a rational point. Hence for given $a,b\in X$ we can certainly find non-parallel $X$-avoiding lines through them.
Hagen von Eitzen
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Let's fix an element in $X$. I am going to choose a point with both coordinates being irrational, for instance $(w,w)\in X$ with $w=\sqrt{2}$.
To prove that $X$ is path connected is enough to prove that for any point $(a,b)\in X$ we can find a path inside $X$ connecting $(a,b)$ to $(w,w)$. Let's prove it
Let $(a,b)\in X$. WLOG say $a$ is irrational. Then go from $(a,b)$ to $(a,w)$ vertically and then go from $(a,w)$ to $(w,w)$ horizontally. Note that this is a path inside $X$ connecting $(a,b)$ and $(w,w)$.
So we are done.
Julian Mejia
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