Closed but not sequentially compact set I read that the intersection of $Q$ (the rational numbers) with $[0,2]$ 'in' the $Q$ field is closed and bounded but not compact. Let this set be called $E$.
The reason is that there can be Cauchy sequence in $E$ that can converge to a number in $R$, such as the square root of 2. But this sequence does not have any subsequence that converges in $E$, hence $E$ is not sequentially compact..
But why is $E$ closed if it does not contain a limit point? (i.e. the square root of 2)... or are limit points required to be defined on the field where the sequence is taken from? ($Q$ in this case)
 A: The set $\mathbb{Q}\cap[0,2]$ is a closed subset of $\mathbb{Q}$ because it can be expressed as the intersection of $\mathbb{Q}$ with a closed subset of $\mathbb R$. And it is true the every limit point of $\mathbb{Q}\cap[0,2]$ $\color{red}{\text{in }\mathbb{Q}}$ belongs to $\mathbb{Q}\cap[0,2]$, which is consistent with the statement that $\mathbb{Q}\cap[0,2]$ is closed $\color{red}{\text{in }\mathbb{Q}}$. So, there is no contradiction here.
A: $\mathbb{Q}\cap[0,2]$ is not closed.
Using the following definition for closed:
"A set $S$ is said to be closed provided that if $\lbrace a_n\rbrace$ is a sequence in S that converges to $a$ then $a$ is also in $S$"
By counterexample, using the series definition of $e$. We can see that if we let 
$a_n=\sum_{k=1}^{k=n}\frac{1}{k!}$ 
then $\lbrace a_n\rbrace\in\mathbb{Q}\cap[0,2]$ and $a_n$ converges to $e-1$ which is not in $\mathbb{Q}\cap[0,2]$. So $\mathbb{Q}\cap[0,2]$ is not closed.

Moreover, a set is closed and bounded if and only if it is compact. So if $\mathbb{Q}\cap[0,2]$ was closed and bounded, then it would have to be compact.
