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!

A subset $C$ of the metric space $(X,d)$ is closed if every sequence $(x_n\in C)$ which converges towards $c$, $c\in C\subset X$.
Here $X=\mathbb{Q}$,, $C=[a,b]\cap\mathbb{Q}$, if you take $a=\sqrt2, b=\sqrt5$ and a sequence of rationals $x_n\in C, limx_n=\sqrt5$, $\sqrt5$ is not in $X$.
What you have is the fact that $C$ is closed in $\mathbb{Q}$ but not in $\mathbb{R}$.
