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