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From the definition, $Y$ is disconnected if $Y=A\cup B,A,B\ne\emptyset A\cap B=\emptyset$ and $A,B$ are both open. So for proving $\mathbb{Q}$ is not connected, normally the textbook would suggest $\mathbb{Q}\cap(-\infty,r)\bigcup\mathbb{Q}\cap(r,+\infty),r\in\mathbb{I}$. Also the textbook would say the open ball of rational number set is not connected. Here's my question.

First, why do we need to consider about $\mathbb{R}$ when proving $\mathbb{Q}$ is not connected?? Aren't we jsut need to consider $\mathbb{Q}$ only? In fact if we just consider $\mathbb{Q}$, it doesn't seem obvious and i first thought that it is connected as i am not sure how to separate it into 2 separated set. Intiutively, if you consider just $\mathbb{Q}$, the only clopen subset in $\mathbb{Q}$ should be itself and $\emptyset$, isn't it? I couldn't find any subset which are clopen in rational set.

Secondly, what if some set which is not familiar with you, how can you really determine whether it is connected or not? Some time the set may be dense and intuitively it seems connected for me but it is not always connected. Some set may even not easy to picture out let alone determine if it is connected

Last, how do one show the open ball $B\subset\mathbb{Q}$ is not connected?

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Connectedness is a topological property, it depends on the topology of $\mathbb{Q}$. You give it the topology it inherits from $\mathbb{R}$ what else do you expect? – Cantor Nov 25 '12 at 16:38
The questioner ought to insist that $A,B$ are both open (or both closed, which implies both open under these conditions). – Ronnie Brown Nov 25 '12 at 16:55
up vote 9 down vote accepted

You don't need to consider $\mathbb{R}$ if you don't want to. For instance, you can take $$A = \{ x \in \mathbb{Q}\, :\, x^2 < 2 \}\ \text{and}\ B = \{ x \in \mathbb{Q}\, :\, x^2 > 2 \}$$ Then $A$ and $B$ are open disjoint sets which cover $\mathbb{Q}$. Both these sets are clopen, since $$ \{ x \in \mathbb{Q}\, :\, x^2 < 2 \} = \{ x \in \mathbb{Q}\, :\, x^2 \le 2 \}$$ or, more concisely written, $\mathbb{Q} \cap (-\sqrt{2},\sqrt{2}) = \mathbb{Q} \cap [-\sqrt{2}, \sqrt{2}]$. In fact, $\mathbb{Q}$ has lots of clopen sets.

The argument that shows that open balls in $\mathbb{Q}$ are disconnected is almost identical.

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I see, for some rational number, there doesn't exist another rational numbers which square of its value equal to it. So there came a separation and we can defined a coninuous 2 valued functions and it is possible to have 2 values so it is not connected – Mathematics Nov 25 '12 at 16:56

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