Prove that if $I$ and $J$ are open intervals and $(I\cap \mathbb{Q}) \cap (J \cap \mathbb{Q}) = \emptyset$ then $I\cap J = \emptyset$.

Question

From first sight I found this problem to be very easy, but then I got stuck! I tried the following:

$(I\cap \mathbb{Q}) \cap (J \cap \mathbb{Q}) = \emptyset$

$I\cap \mathbb{Q} \cap J \cap \mathbb{Q} = \emptyset$

$I \cap J \cap \mathbb{Q} = \emptyset$

If we introduce $\mathbb{I}$ such that $\mathbb{I}$ is the set of irrationals we get:

$I \cap J \cap \mathbb{Q} \cap \mathbb{I} = \emptyset \cap \mathbb{I}$

It here when realized that this gives me : $I \cap J \cap \emptyset = \emptyset \cap \mathbb{I}$ $\implies \emptyset = \emptyset$

Edit: I and J are open intervals.

Answer 1

Let $I, J$ be open intervals and suppose $(I\cap \mathbb{Q}) \cap (J \cap \mathbb{Q}) = \emptyset$.

We show that $I \cap J = \emptyset$ by contradiction. Suppose, for contradiction, that there is $x \in I\cap J$. Since $I$ (resp. $J$) is open, there is $\epsilon' > 0$ (resp. $\epsilon''$) such that the open neighbourhoud $]x - \epsilon' ; x + \epsilon' [$ is contained in $I$
(resp. $]x - \epsilon'' ; x + \epsilon'' [ \subset J$).

Now set $\epsilon := \min(\epsilon', \epsilon'')$. The open neighbourhoud $O :=]x - \epsilon ; x + \epsilon [$ is contained in $I \cap J$. Since every non-empty open interval of the reals contains an element from $\mathbb{Q}$, we get $O \cap \mathbb{Q} \neq \emptyset$ whence $I \cap J \cap \mathbb{Q} \neq \emptyset$, a contradiction.

Answer 2

Actually, if $I=J=\{\pi\}$, then $(I\cap\Bbb Q)\cap(J\cap\Bbb Q)=\emptyset$, but $I\cap J=\{\pi\}$.

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If $I$ and $J$ are non-empty open intervals, then the statement is true. Indeed, if $I\cap J\ne\emptyset$, then $I\cap J$ is also an open non-empty interval. Therefore, it contains some ratinal number $q$. That is, $q\in(I\cap J)\cap\Bbb Q$. But$$(I\cap J)\cap\Bbb Q=(I\cap\Bbb Q)\cap(J\cap\Bbb Q).$$