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: To say that a subset $K \subset \mathbb Q$ is closed in $\mathbb Q$ means that for every sequence $(x_i)$ in $K$, if $(x_i)$ converges to a limit $L$ which is in $\mathbb Q$ then $L$ is in $K$.
Now, you have found a sequence $(x_i)$ in $K$ that converges to a limit $L$ which is in $\mathbb R$ and is not in $K$. However, this is not a contradiction to the statement that $K$ is closed in $\mathbb Q$. To find a contradiction, you would have needed to find a sequence $(x_i)$ in $K$ that converges to a limit $L$ which is in $\mathbb Q$ and is not in $K$, and you have not done that.
A: 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 $\mathbb{R}$.
A: 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:
$$K = \underbrace{[a,b]}_{\text{closed in }\Bbb{R}} \cap \Bbb{Q}.$$
