Real Analysis -Is ℚ ∩ [0,1) sequentially compact?

Question

So I've been given this question that asked whether ℚ ∩ [0,1)is sequentially compact, and I think it isn't due to the fact that it is neither bounded nor closed, but I am basing that off of instinct, so what is a formal way of justifying this?

Answer 1

To show directly that your set does not satisfy the definition of "sequentially compact", you would give an example of a sequence that has subsequence that has a limit within the set.

For example, your sequence could be $$ 0, \frac12, \frac23, \frac34, \frac45, \ldots, \frac{n-1}{n}, \ldots $$ Can you show that no subsequence of this has a limit in $\mathbb Q\cap[0,1)$?

(Hint: In $\mathbb R$, $1$ is a limit of every possible subsequence because ...? This implies ... because ...?)

Answer 2

Since $\Bbb Q$ is dense in $\Bbb R$, take a sequence $(x_n)$ in $\Bbb Q\cap[0,1)$ converging to, say, $\sqrt2-1$. Then every subsequence converges to a point not in $\Bbb Q\cap[0,1)$.