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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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    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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    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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    sorry it is wrong2016-08-19