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This question refers to How to prove closure of $\mathbb{Q}$ is $\mathbb{R}$

I want to prove that the closure of $\mathbb{Q}$ is $\mathbb{R}$. I am trying to understand the accepted answer, but when it comes to "This shows that the complement of $\mathbb{Q}$ has empty interior, so the closure of $\mathbb{Q}$ is all of $\mathbb{R}$.", I get stuck.

What does it mean intuitively that a set has empty interior?

What I know is that a point is interior to a set when it is the center of some open ball inside that set. In that sense, for a set to have empty interior would mean that it has no interior points. Furthermore, the set of all its adherent points - a point $m$ is adherent to the set of rationals when there is a sequence of rational points $x_k$ such that $limx_k = m.$ Being so, the complement of rationals would have no interior points.

How does it follow from this that the closure of $\mathbb{Q}$ is $\mathbb{R}$?

Thanks in advance!

Community
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DrHAL
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5 Answers5

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How does it follow from this that the closure of $\mathbb{Q}$ is $\mathbb{R}$?

Take $x \in \mathbb{R}$. Since $\mathbb{R}\ - \mathbb{Q}$ has empty interior, it follows that $x \notin int(\mathbb{R} - \mathbb{Q})$. It means that, for all $r > 0$, the ball $B(x,r)$ centred in $x$ with radius $r$ is not contained in $\mathbb{R}\ - \mathbb{Q}$. So, there is a $q \in \mathbb{Q}$ such that $q \in B(x,r)$. In other words, $x$ is in the closure of $\mathbb{Q}$. Since $x$ is arbritary, $\overline{\mathbb{Q}} = \mathbb{R}$.

remath7
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  • I understood that last part of the question! Now, how can I prove that every open ball around a real number r contains a rational is equivalent to $\mathbb{R-Q}$ having empty interior? @remath7 – DrHAL Apr 27 '16 at 04:15
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    @DrHAL It's simple. Suppose $\mathbb{R} - \mathbb{Q}$ have non empty interior. So, there existe a open ball $B \subset \mathbb{R} - \mathbb{Q}$. Hence, $B$ cannot contain a rational number, but it's a contradiction, since every open ball contains a rational number. – remath7 Apr 27 '16 at 04:42
  • Then I would still have to prove that every open ball contains a rational number, right? @remath7 – DrHAL Apr 27 '16 at 05:05
  • Yeah, but I have shown that it's equivalent to $\mathbb{R} - \mathbb{Q}$ to have empty interior. Since, as you said, it was already proved, the result follows. Your question was why these two fact are equivalent. – remath7 Apr 27 '16 at 12:38
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This depends largely on the topology of the space that you're in. As copper.hat noted above, a set in a topological space has empty interior if it contains no open set. Recall that on $\mathbb{R}$ the open sets are generated by open intervals of the form $(a,b) = \{x \in \mathbb{R} \; | \; a <x < b\}$. So the complement of the rationals, which we can denote $\mathbb{R}\setminus \mathbb{Q}$, contains no interval. You can show this by demonstrating that a ball around any irraitonal point must contain a rational element.

Mnifldz
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Every real number can be approximated arbitrarily closely by rational numbers; this is the basis for decimal notation. Now for an open set contained in the complement of $\Bbb Q$, its elements by definition cannot be approximated arbitrarily closely by rational numbers. So such a set is empty QED.

Marc van Leeuwen
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In any topological space $X$, for any subset $A$ of $X$, the interior of $A$ is the union of all open subsets contained in $A$, and the closure of $A$ is the intersection of all closed subsets that contain $A$. This is not the definition of interior and closure, but these characterizations follow easily from the definitions. (And each of the two propositions follows from the other, too, by taking complements). In particular, it follows that $A$ is dense in $X$ if and only if the complement of $A$ has empty interior. (The first condition means that $X$ is the only closed subset containing $A$; by looking at complements, this is the same as: the only open subset contained in the complement of $A$ is the empty set.)

Note that all this holds in any topological space, not only in metric spaces. There are no hypotheses required on $X$ or on $A$.

  • Can you elaborate the part where you say A is dense in X if and only if the complement of A has empty interior? I am particularly trying to understand how the complement of rationals (irrationals) have empty interior following from the part where every ball and how that is related to the closure... I am very confused. @mathguy – DrHAL Apr 27 '16 at 04:15
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a neighborhood of a point $x$ in $\mathbb{R}$ is an open interval, $(x- \epsilon _1 , x+ \epsilon _2 )$.

if $x$ is an interior point of a set $S \subseteq \mathbb{R}$, then x has a neighborhood of that flavor, which is contained in $S$.

if a set doesn't have any interior points, then, given any point in the set, you can't find an open interval $(x-\epsilon , x+\epsilon)$ which is contained in the set

in other words, if a subset of $\mathbb{R}$ has empty interior, then it doesn't have any subsets which are intervals

homedoggieo
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