# Is $\mathbb Q^n$ disconnected?

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. –  Cameron Buie Sep 20 '12 at 2:15
What is $i$? Do you mean that $p_1, p_2, \ldots p_n$ all have to be rational? –  Chris Eagle Sep 20 '12 at 2:21
Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. –  Martin Sleziak Sep 20 '12 at 16:43
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. –  Mariano Suárez-Alvarez Sep 20 '12 at 16:45

HINT: Consider the set $\{\langle p_1,p_2,\dots,p_n\rangle\in X:p_1<\sqrt2\}$. Is it open? Closed?
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. –  Alex Youcis Sep 20 '12 at 2:28