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there is a problem on topology.

Let $n > 1$ and let $X = \{(p_1,p_2, \ldots , p_n)\mid p_i\text{ is rational}\}$. Show that $X$ is disconnected.

how to solve this problem.i am completely stuck out.

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    By disconnected, do you mean "not connected", or "totally disconnected"? Both are true, but you'll need to approach it differently, depending on which you intend.2012-09-20
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    What is $i$? Do you mean that $p_1, p_2, \ldots p_n$ all have to be rational?2012-09-20
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    Please, try to make the titles of your questions more informative. E.g., *Why does $a2012-09-20
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    If the space $X$ is diconnected, then for all spaces $Y$ the product $X\times Y$ is also disconnected: this implies that it is enough to deal with the case $n=1$ of the problem.2012-09-20
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    What exactly are you stuck on? Start with the definition of connectedness (or totally disconnectedness), and look for results that would give you the conclusion you want. You can't be completely stuck on something that's this close to the definitions involved.2016-08-18

4 Answers 4

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HINT: Consider the set $\{\langle p_1,p_2,\dots,p_n\rangle\in X:p_1<\sqrt2\}$. Is it open? Closed?

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    To add, you can reduce to this to merely checking $\mathbb{Q}$, since if $\mathbb{Q}^n$ were connected then so would $\mathbb{Q}$, being the image of $\mathbb{Q}^n$ under the continuous projection map.2012-09-20
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I'll show it's totally disconnected, i.e. if $S \subseteq Z$ contains two elements or more, then it can be written $ S = A \cup B $ where $A, B$ are open, disjoint, and have nonempty intersection with $S$.

Since $S$ contains at least two distinct elements, say $\mathbf{p}, \mathbf{q}$, then for some $j \in \{ 1, \ldots, n \}$, we have $p_j \neq q_j$. Suppose WLOG that $p_j < q_j$. Then there exists an irrational number $\theta \in ( p_j , q_j )$. Let $A = \{ \left< a_1 , \ldots , a_n \right> \in \mathbb{Q}^{n} : a_j < \theta \} , B = \{ \left< a_1 , \ldots , a_n \right> \in \mathbb{Q}^{n} : a_j > \theta \}$. Then $A, B$ are open in $\mathbb{Q}^n$, and $\mathbf{p} \in A , \mathbf{q} \in B$. Moreover, $A, B$ are disjoint, open, so $S$ is not connected.

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Assume $\mathbb{Q}^n$ is connected. Let $\pi : \mathbb{Q}^n \to \mathbb{Q}$ be a projection map with usual topology. As $\pi$ is continuous $\mathbb{Q}$ will be connected, which is a contradiction. Therefore $\mathbb{Q}^n$ is disconnected.

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    Is that proof wrong? It looks correct to me (even if the post could be written in a better way).2016-08-18
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    it is just in short form2016-08-19
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If we take the line $Y=(\sqrt 2)Z + (\sqrt 3,\sqrt 3,\sqrt 3, \cdots \text{n times})$ where $Y,Z\in R^n$. then the line cut the space into two disjoint open sets.hence it is disconnected

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    How does a _line_ cut the _space_ into parts?2016-08-17
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    sorry it is wrong2016-08-19