Questions on proving that every infinite set of real numbers has a countable dense subset

Question

I have been learning answers provided in this question. The answer from Fishbane uses this definition of a dense subset.

A subset $B$ of $A$ is dense in $A$ if for every $\varepsilon > 0$ and any $x \in A$, there exists an element $y \in B$ such that $|x-y|<\varepsilon$.

I am currently using the book of Strichartz, "The Way of Analysis" (revised edition) in which a dense subset is defined as follows

A subset $B$ of $A$ is dense in $A$ if $A$ is contained in the closure of $B$, i.e. $B \subseteq A \subseteq \bar{B}$.

To prevent any difference between definitions, I will also provide the definitions of open sets, closed sets, limit points of a set, and closure. I will put these definitions in this image so that this question is not too long.

I have proved that both definitions of dense subsets are equivalent, i.e if I am following the definition in Strichartz, then

A subset $B$ of $A$ is dense in $A$ if and only if for every $\varepsilon > 0$ and any $x \in A$, there exists an element $y \in B$ such that $|x-y|<\varepsilon$.

proof. $(\Rightarrow)$ Let $B$ be a dense subset of $A$. Then, $B \subseteq A \subseteq \bar{B}$. Let $x$ be an arbitrary element of $A$. Thus, either $x \in B$ or $x$ is a limit point of $B$. If $x \in B$, then choose $y = x$ and we are done. If $x$ is a limit point of $B$, then $\forall n \in \mathbb{N}, \exists y_n \neq x \ni |x-y_n|<\frac{1}{n}$. By Archimedian property, this immediately applies for every $\varepsilon > 0$.

$(\Leftarrow)$ Let $A$ and the subset $B$ of $A$ have the property that for every $\varepsilon > 0$ and any $x \in A$, there exists an element $y \in B$ such that $|x-y|<\varepsilon$. Let $x \in A$ be arbitrary, then either $x \in B$ or $x \in A\backslash B$. It is trivial if $x \in B$, so that $x \in \bar{B}$. If $x \in A\backslash B$, then $\forall n \in \mathbb{N}, \exists y_n \in B \ni |x-y_n|<\frac{1}{n}$ as $\frac{1}{n} > 0$ for all $n \in \mathbb{N}$. Since $x \in A\backslash B$, it must be $y_n \neq x$ for all $n$, hence $x$ is a limit point of $B$, implying that $x \in \bar{B}$.

Now I will provide my proof that every infinite set of real numbers has a countable dense subset using the equivalence between definitions. Let $A$ be an infinite set and $n \in \mathbb{N}$. This proof is actually the detailed version of Fishbane's answer, so I immediately start with the construction of $B_n \subseteq A$ with property that $B_n$ is countable and $\forall n \in \mathbb{N}, \exists y \in B_n \ni |x-y|<\frac{1}{n}$. Let $Z_n$ be the set defined below $$Z_n = \left\{ \left[\dfrac{k}{n},\dfrac{k+1}{n}\right) \cap A : k \in \mathbb{Z} \right\} \backslash \{\emptyset\}.$$ Surely there are infinitely (countably to be precise) many integers $k$ such that $\left[\dfrac{k}{n},\dfrac{k+1}{n}\right) \cap A$ are nonempty since $A$ is infinite. Therefore, by axiom of countable choice, we can construct a countable set $B_n$ from $Z_n$, say $$B_n = \{ x_n^{(k_1)}, x_n^{(k_2)}, x_n^{(k_3)}, ... \}.$$ Now, since $x \in A$, then $nx \in \mathbb{R}$. Therefore, there is an integer $k'$ such that $$ k' \leq nx < k'+1 \Longrightarrow \dfrac{k'}{n} < x <\dfrac{k'+ 1}{n}.$$ This imply that $\left[\dfrac{k'}{n},\dfrac{k'+1}{n}\right) \cap A$ is nonempty, hence there exists $x_n^{(k')} \in B_n$ such that $x_n^{(k')} \in \left[\dfrac{k'}{n},\dfrac{k'+1}{n}\right) \cap A$. Therefore, $|x - x_n^{(k')}|<\frac{1}{n}.$

Continuing for every natural number $n$, we have $B = \bigcup\limits_{n=1}^{\infty} B_n$, which is countable since it is a countable union of countable sets, that satisfies the property

For every $n \in \mathbb{N}$ and any $x \in A$, there exists an element $y \in B$ such that $|x-y|<\frac{1}{n}$. By Archimedian property, this immediately applies for every $\varepsilon > 0$.

Therefore, we obtain a countable subset $B$ of $A$ that satisfies the desired property, so it is proved that every infinite set of real numbers has a countable dense subset.

My questions:

Is my proof correct? The part that bothers me the most is that the implication of $\left[\dfrac{k'}{n},\dfrac{k'+1}{n}\right) \cap A$ being nonempty after we know that $x \in \left[\dfrac{k'}{n},\dfrac{k'+1}{n}\right)$. Do we have to guarantee that the element $x_n^{(k')}$ differs from $x$ if $\left[\dfrac{k'}{n},\dfrac{k'+1}{n}\right) \cap A$ contains more than one point? even though, the equivalence states that it is sufficient to choose an element in $B$ without any restriction that it has to differ from $x$ on certain condition.

I also tried the hint provided by Mariano Suárez-Álvarez. He suggests another approach to constructing the countable subset $B$ of $A$ using nonempty intersection of rational intervals and $A$. This also leads to the same problem when implying that some $(p,q) \cap A$ is nonempty, hence take $x_{p,q}$ so that $|x - x_{p,q}| < \frac{1}{n}$.

Any clarification or insight would be appreciated.

Answer 1

Your proof looks fine overall. A few points:

Surely there are infinitely (countably to be precise) many integers $k$ such that $\left[ \frac{k}{n}, \frac{k + 1}{n} \right) \cap A$ are nonempty since $A$ is infinite.

This isn't true. Consider for instance $A = [0, 1]$, which is infinite. For any given $n \in \mathbb{Z}^+$, the number of such intervals that intersects $A$ is $n + 1$. Importantly, it is finite: countable, but not infinite.

This is a small point and doesn't really affect the rest of your proof.

Do we have to guarantee that the element $x_n^{(k')}$ differs from $x$ if $\left[ \frac{k'}{n}, \frac{k' + 1}{n} \right) \cap A$ contains more than one point?

It's perfectly fine for $x$ not to differ from $x_n^{(k')}$, even if the intersection contains more than one point. Let's take your definition of "dense subset" from Strichartz:

A subset $B \subseteq A$ is dense in $A$ if $A$ is contained in the closure of $B$; i.e., $B \subseteq A \subseteq {\rm Cl} B$.

Letting $x \in A$ be arbitrary, we just require that for any $\varepsilon > 0$, the neighborhood $B_d (x, \varepsilon)$ of $x$ intersects $B$. If we get $x \in B$, all the better!