If $a,b$ are irrational numbers, Is $K=[a,b] \cap \mathbb Q$ closed in $\mathbb Q$? Suppose $a,b \in \mathbb R-\mathbb Q , a <b$. Consider $K=[a,b] \bigcap \mathbb Q$.
Now, $[a,b]$ is closed in the metric superspace of $\mathbb Q$ i.e in $\mathbb R$. Thus, $K=[a,b] \bigcap Q$ is closed in $\mathbb Q$
But, $K=[a,b] \bigcap Q=(a,b) \bigcap Q$ doesn't contain it's limit points, namely, $a,b$ . Suppose $a=\sqrt 2, b = \sqrt 5$. 
If we define a function $A= \{x \in K~|~x^2 <5 \}$, then there is a sequence in $K$ which converges to $\sqrt 5$ but $\sqrt 5 \notin K .$
Then how can K be closed in $\mathbb Q$?
Thanks a lot for the help!

Since $\Bbb{Q}$ inherits the topology of $\Bbb{R}$, $K$ is closed in $\Bbb{Q}$ if and only if we can write it as an intersection of a closed set in $\Bbb{R}$ with $\Bbb{Q}$. But this is true:
$$K = \underbrace{[a,b]}_{\text{closed in }\Bbb{R}} \cap \Bbb{Q}.$$
