$\mathbb{Q}$ is closed in itself It is known that every space is both open and closed in itself. If the space is a metric vector space, then being closed is equivalent that every sequence that converges, converges to a point in the space.
For example, $\mathbb{Q}$ is closed in itself. What does it say about the sequence of rational approximations of some irrational number? Does it not converge? It is a Cauchy sequence so how can we say it does not converge?
 A: I think you are mixing two concpets:
Yes, $\mathbb Q$ is a "closed set", since $(\Omega, \mathscr T)$ as a topological space by its own is always defined as open and closed, or we sometimes call it clopen.
But at the same time, we call a space "complete" if every Cauchy sequence converges within the space - and in this case, $\mathbb Q$ is not complete.
A: $(\mathbb{Q},|.|)$ is closed in itself.
But a sequence of $q_n$ rational numbers which converges to an irrational $r$, does not converge in $\mathbb{Q}$
It is a Cauchy sequence but does not converge in $(\mathbb{Q},|.|)$
Thus we can say that $\mathbb{Q}$ with the subspace topology inherited from the usual topology of $\mathbb{R}$ is not a complete space.
Take for instance $x_n=(1+1/n)^n \rightarrow e$.
Thus $x_n$ is a Cauchy sequence but $e \notin \mathbb{Q}$
A: If $A$ is a subset of a topological space $X$, then "$A$ is closed (in $X$)" implies that if a sequence of points of $A$ converges to a point in $X$, then the limit is itself in $A$.
This doesn't say anything about the possibility of converging to points outside $X$ -- indeed, the topology on $X$ that declares $A$ to be closed does not even know whether such points exist or what it would mean to converge towards them.

What does it say about the sequence of rational approximations of some irrational number? Does it not converge? It is a Cauchy sequence so how can we say it does not converge?

It doesn't converge in $\mathbb Q$ because there is no point in the topological space that it converges to. Some metric spaces have the property that all Cauchy sequences converge (they're called complete), but $\mathbb Q$ is not one of them.
A: The "Cauchy criterion" does NOT hold in the rational numbers.  There are, as you say, "Cauchy" sequences of rational numbers that do not converge to a rational number.
That is, in fact, one way of "constructing" the real number system:  We say that two Cauchy sequences of rational numbers are "equivalent" if and only if their "difference" (if the sequences are $\{a_n\}$ and $\{b_n\}$ then their "difference" is the sequence $\{a_n- b_n\}$.)  converges to 0.  We then identify the equivalence classes with the real numbers.  It is easy to show that if two Cauchy sequences of rational number converge to the same rational number then they are equivalent in this sense and that any other such sequence that is equivalent converges to the same rational number.  We identify that equivalence class with that rational number so that, in this sense, the set of rational number is a subset of the set of real number.  But since there exist Cauchy sequences that do not converge to a rational number, the corresponding equivalence classes give us the irrational numbers.
A: The equivalence statement you stated in the first paragraph is more tricky than it appears! Note that the necessary and sufficient condition you stated is equivalent to that the set equals its closure. However, when we say $\mathbb{Q}$ is closed in itself, we are certainly not saying $\mathbb{Q}$ is its closure that is usually taken to be $\mathbb{R}$ for granted unless otherwise declared. You probably are having trouble with the "inconsistency" of these pieces of information. The crux is this. We say $\mathbb{Q}$ is closed in itself because $\mathbb{Q}^{c} = \varnothing$ is open by definition. But note that here $\mathbb{Q}^{c}$ is the complement of $\mathbb{Q}$ with respect to $\mathbb{Q}$. So, when $\mathbb{Q}$ itself is considered the ambient space, the closure of $\mathbb{Q}$, which is by definition (if you are not using this definition, never mind; they are equivalent.) the intersection of all closed sets that includes $\mathbb{Q}$, is $\mathbb{Q}$ itself. So, if you go through the proof of the equivalence theorem again, then you will see why the "contradiction" resolves. 
