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Can somebody explain me why the following set in neither open nor closed.

$$ B:=\{ (x,y)\in \mathbb Q^2: 1\le x,y\le 10 \}$$

My thoughts:- If I can find an $\epsilon$ neighborhood that belongs to the set at every point in the set then it is open. If the complement set of the set is open then the set is closed.

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    It might be easier to work with the sequential definitions. – 2017-02-16
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    $B$ is a closed subset of $\Bbb Q^2$, but not of $\Bbb R^2$. – 2017-02-16

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Let $(x,y) \in B$ and $U$ a $ \epsilon$ - neighborhood of $(x,y)$. Then $U$ contains infinitely many points $(a,b)$ with $(a,b) \notin \mathbb Q^2$. So, no $ \epsilon$ - neighborhood of $(x,y)$ is contained in $B$. Therefore $B$ is not open.

We have $ \overline{B}=\{ (x,y)\in \mathbb R^2: 1\le x,y\le 10 \}$, hence $B \ne \overline{B}$ and $B$ is not closed.