# Closure of the Irrational [closed]

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.

## closed as unclear what you're asking by Mohammad Riazi-Kermani, hardmath, Shailesh, The Phenotype, Arnaud MortierMar 11 '18 at 17:21

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• You can indeed use the complementary set of $\Bbb R - \Bbb Q$, and the fact that the interior of this set is .... ? – krirkrirk Mar 11 '18 at 14:36
• it's the very $\Bbb R$ – Muradin Bronzebeard Mar 11 '18 at 14:38
• No, I was talking about the interior of $\Bbb Q$, not its closure. – krirkrirk Mar 11 '18 at 14:41
• The interior of $\Bbb R - \Bbb Q$ is the complement of the closure of the complement of $\Bbb R - \Bbb Q$? – Muradin Bronzebeard Mar 11 '18 at 14:44
• Once again, I'm talking about the interior of $\Bbb Q$, not $\Bbb R - \Bbb Q$. – krirkrirk Mar 11 '18 at 14:46

## 2 Answers

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 $\overline{\mathbb{I}}=\mathbb{R}$

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

1. 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 $\overline{X} = (\text{int}({X^c}))^c$.