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How to prove: $Cl_{\mathbb{R}}(\mathbb{IQ})=\mathbb{R}$ ($\mathbb{IQ}$-set of irrational numbers)?

I know how to prove: $Cl_{\mathbb{R}}(\mathbb{Q})=\mathbb{R}$ ($\ast$)

Proof of ($\ast$): Let: $r \in \mathbb{R}$ , $q \in \mathbb{Q}$, $\epsilon \in \mathbb{R}_{+}$. Then: $\exists n \in \mathbb{N}:$ $\frac{1}{10^{n}}<\epsilon$. Write $r$ in decimal expansion. Let q have the same decimal expansion up to $n-th$ place. Then: $\mid r-q\mid<\frac{1}{10^{n}}<\epsilon$. This implies: every ball $B(r,\epsilon)$ contains $q\in \mathbb{Q}$ $\square$

Thought of constructing some analogical proof, however I failed.

I would be very thankful for help.

  • Use the fact that irrationals are dense in $\mathbb R$ – windircurse Nov 17 '17 at 19:28
  • I know this. However, I do not see how to construct formal proof. –  Nov 17 '17 at 19:30
  • I know that, intuitively speaking, for every neighbourhood of real number exists irrational belonging to it. –  Nov 17 '17 at 19:31
  • The problem is how to formulate it rigorously. –  Nov 17 '17 at 19:31
  • I would argue that there are issues with your 'proof': how do you know that you can write $r$ with a decimal expansion? What is a decimal expansion? (speaking formally - informally we all know what it is, but a lot of the point of these exercises is to understand how to work with formal definitions, not informal ones.) The correct way of thinking about this problem may depend a lot on your specific definition of $\mathbb{R}$. – Steven Stadnicki Nov 17 '17 at 19:47
  • https://math.stackexchange.com/q/935808/169852 –  Nov 17 '17 at 22:57
  • Thanks. This post helps. –  Nov 19 '17 at 12:12

3 Answers3

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The truncating decimal argument you gave is one fine way to see it. In a way it follows from the definition of the reals, really.

Another way is to prove $\sqrt{2}$ is irrational and appoximate reals with numbers like $q + r\sqrt{2}$ which are all irrational, if $q,r \in \mathbb{Q}\setminus \{0\}$ etc. Or $q + \frac{1}{n}\sqrt{2} -> q, (n \to \infty)$ so $\mathbb{Q} \subseteq \overline{\mathbb{P}}$ and hence $\mathbb{R}=\overline{Q} \subseteq \overline{P}$ etc.

Henno Brandsma
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  • Or use the density of $\Bbb Q$ in $\Bbb R.$ If $x,y\in \Bbb R$ with $x<y,$ let $x'\in \Bbb Q \cap (x,y)$ and let $y'\in \Bbb Q\cap (x',y).$ Then $x'+(y'-x')/\sqrt 2;\in (x,y)$ \ $\Bbb Q.$ Although we can replace $\sqrt 2$ with any irrational that's greater than $1.$ – DanielWainfleet Nov 19 '17 at 21:58
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I think in this case the most advantageous approach is to first thoroughly understand the density of $\mathbb{Q}$ in $\mathbb{R}$.

Thus the density of $\mathbb{IQ}$ in $\mathbb{R}$ follows with little to no effort from the density of $\mathbb{Q}$ in $\mathbb{R}$.

Then $ Cl_{\mathbb{R}}(\mathbb{IQ}) = \mathbb{R} $ follows immediately just by the very definitions of closure and density.

Because by the definition of closure, $ Cl_{\mathbb{R}}(\mathbb{IQ}) \subseteq \mathbb{R} $, and by the definition of density $ \mathbb{R} \subseteq Cl_{\mathbb{R}}(\mathbb{IQ}) $.

Therefore $ Cl_{\mathbb{R}}(\mathbb{IQ}) = \mathbb{R} $.

mucciolo
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My solution:

Lemma: $\mathbb{Q}$ dense in $\mathbb{R}$

Proof(of lemma): Let: $x,y \in \mathbb{R}:$ $x<y$, Density of $\mathbb{Q}$ implies: $\exists q \in \mathbb{Q}$: $q \in(x,y)$. $\frac{y-r}{2}>0$ $\implies$ Archimedean axiom implies: $\exists n \in \mathbb{N}$: $\frac{y-r}{2}>\frac{1}{n}$ $\implies$ $y > r + \frac{2}{n} > r+ \frac{\sqrt{2}}{n}>r+0>x$ $\implies$ $\exists s:=r+\frac{\sqrt{2}}{n}\in\mathbb{IQ}$: $r \in(x,y)$ $\square$

Lemma implies: $\forall r \in \mathbb{R}_{+}$ $\forall x \in \mathbb{R}$: $B(x,r) \cap \mathbb{IQ} \neq \emptyset$ $\implies$ $\mathbb{R}=\{x \in \mathbb{R} \mid \forall U_{x}: U_{x}\cap\mathbb{IQ}\neq\emptyset\} = Cl_{\mathbb{R}}(\mathbb{IQ})$