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Do all subsets $A \subset \mathbb{R}$ have a countable dense subset $B \subset A$? I first thought about $A \cap \mathbb{Q}$ but if $A= \mathbb{R} \setminus \mathbb{Q}$ this doesn't work

mathlander
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3 Answers3

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There is a simple construction that can be used. We shall use the definition of dense in $A$ that for any $\epsilon>0$ and any element $x \in A$ there is an element $y \in B$ such that $|x-y|<\epsilon$.

Now let us construct sets $B_n \subseteq A$ with the property that $B_n$ is countable and for each element $x \in A$ there is an element $y \in B_n$ such that $|x-y|<\frac{1}{n}$.

To construct $B_n$ we can use the axiom of choice to choose one element from each of the nonempty sets $[\frac{k}{n},\frac{k+1}{n}) \cap A$ (for $k \in \Bbb{Z}$) and let $B_n$ be the collection of all of these. Clearly $B_n$ satisfies the required properties.

Now it is clear that $B = \bigcup\limits_{n=1}^{\infty} B_n$ is countable and satisfies the definition of a dense set in $A$ and so we are done. $\square$

Hope this helps. If there are any issues let me know.

Fishbane
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  • I'm not sure I understand this answer. When you write "let $B_n$ be the collection of all of these", does $k$ run over all of the integers, or just the natural numbers? Also, how do you know that $[k/n,(k+1)/n]\cap A$ is nonempty? Certainly it can be empty for some values of $k$, $n$, and sets $A$. What if $A=\varnothing$, for instance? – Joe Jan 25 '23 at 17:06
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    @Joe Yes $k$ does indeed run over all the integers, I should have made that more clear. We do not know that $[\frac{k}{n},\frac{k+1}{n}) \cap A $ is not empty in fact that is why I say "each of the nonempty sets $[\frac{k}{n},\frac{k+1}{n}) \cap A$". What I mean is construct those sets and then discard any that are empty. i.e. use the axiom of choice on the following set ${[\frac{k}{n},\frac{k+1}{n}) \cap A | k \in \Bbb{Z} \text{ and } [\frac{k}{n},\frac{k+1}{n}) \cap A \neq \emptyset}$. – Fishbane Jan 25 '23 at 17:15
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    Got it, thanks. Regarding your second point, I read what you wrote to mean "each of these sets is nonempty, so we can choose one element from them". I now understand what you intended. – Joe Jan 25 '23 at 17:22
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    Incidentally, we also have to use a fragment of the axiom of choice to conclude that a countable union of countable sets is countable. For readers who are interested, see Peter Smith's answer to this question. – Joe Jan 25 '23 at 17:23
  • @Joe I agree it was perhaps poorly worded. Thanks. Wow, I didn't realize we needed the axiom of choice for that but now that you say it it does feel like it makes sense. – Fishbane Jan 25 '23 at 17:23
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    Yes, it is quite unexpected! As I understand, there are models of $\mathsf{ZF}$ where the real numbers themselves are a countable union of countable sets, meaning that the theory of Lebesgue measure fails completely. This shows the importance of the axiom of choice in analysis – or at least, weak forms of the axiom of choice. This is discussed by Joel David Hamkins in this thread. – Joe Jan 25 '23 at 17:41
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This follows from subspaces of separable metric spaces are separable.

For $\Bbb R$ is a separable metric space. Note this is not true in general (non-metric) topological spaces (for instance the Sorgenfrey plane).

calc ll
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For each pair of rational numbers $a$ and $b$ such that $a<b$ and $(a,b)\cap A\neq\emptyset$ let $x_{a,b}$ be any point of that intersection.

The set of all those points is countable and dense in $A$.

Mariano Suárez-Álvarez
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