Closure of the Irrational

Question

I have to find the closure of the Irrational in the context of Euclidean topological space. I would have to find your complementary set, right? And it should be closed in R.

Can you help me? I can not prove it.

Answer 1

As you know $\mathbb{R}=\mathbb{Q}\cup\mathbb{I}$ is a disjoint union, the irrationals are dense in the real numbers, (also the rationals),i.e. $\forall{x,y}\in\mathbb{R}$ with $x>y,\exists{I}\in\mathbb{I}$ (the irrationals) such that $x>i>y$. For this, the border points of $\mathbb{I}$ are $\mathbb{Q}$ and $\mathbb{I}$, and the interior points of $\mathbb{I}$ are the vacuum. You know the clausure of a set is its interior union its border, then you have $\overline{\mathbb{I}}=\mathbb{R}$

Answer 2

Following our discussion in the comments : one way (amongst many) to prove the density of the irrationals is as follow :

• Prove that $\mathring{\Bbb Q} = \emptyset$
• Use the following to conclude :

  $X$ is dense in $\Bbb R$ $\iff \Bbb R -X$ has empty interior

Let's prove both those points.

1. $$\begin{align} \mathring{\Bbb Q} = \emptyset & \iff \text{ any open subset of $\Bbb Q$ contains an irrational number} \\ &\iff \forall q \in \Bbb Q, \forall \epsilon > 0, B_\epsilon(q) = \{ x\in \Bbb R\ : \vert x - q \vert < \epsilon \} \text{contains an irrational number.} \end{align}$$

But any interval $]a;b[$ contains an irrational number, because :

• if there is no irrational in this interval, then $\Bbb R \subseteq \Bbb Q$, since $\forall x \in \Bbb R$, $\exists r \in \Bbb Q$ s.t. $x+r\in ]a,b[$. - left as an exercice
• but $\Bbb R \subseteq \Bbb Q$ is obviously false - why ?

This shows $\mathring{\Bbb Q} =\emptyset$.

2. Let's prove :

   $X$ is dense in $\Bbb R$ $\iff \Bbb R -X$ has empty interior

Actually, we're only interested in $\impliedby$. You should be able to prove this, using the fact that $\overline{X} = (\text{int}({X^c}))^c$.