Consider the space $\mathbb R$, as you may know it is a complete metric space. This means that every bounded monotonic sequence is convergent.
For example, $a_n=\dfrac1n$ converges to $0$.
If, however, we consider only a subspace of the real numbers, for example $(0,1)$ - all the numbers strictly between $0$ and $1$ then the above sequence does not converge in that space. Why? For a sequence to be convergent the space has to know the possible limit point. The limit is $0$ which $(0,1)$ does not know about.
If you take the number $\sqrt 2$, as you may know it is not a rational number. Take the rational sequence defined by longer and longer decimal expansions of $\sqrt 2$: $1,1.4,1.41,\ldots$
In the real numbers this is a monotonic and bounded sequence (all those are below $\sqrt 2$) and thus converge (to $\sqrt 2$ as luck would have it). Now you can see that all those elements in the sequence are rationals - the decimal expansion is finite - therefore this sequence is also a sequence in $\mathbb Q$.
In the rational numbers this is a monotonic and bounded sequence but the limit point is $\sqrt 2$ and this is a number that $\mathbb Q$ does not know about and therefore cannot tell that the sequence is converging.