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$F \subset \mathbb{R}$ is closed. Prove there exists a countable $E \subset F$, where $\bar{E} = F$. ($\bar E$ means closure of $E$)


proof 1

Let $E$ be the interior of $F$, which is open, by definition. Clearly, $\bar E = F$. Since every open set in $\mathbb{R}$ can be written uniquely as the countable union of disjoint open intervals, and open intervals are opens sets, the set $E$ must be countable.

Note: That seems way too easy.

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    $\begingroup$ You are correct in being suspicious. $\endgroup$ – Randall Sep 25 '18 at 1:05
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    $\begingroup$ The sentence starting with "Clearly" is false. Consider the Cantor set, which is closed but has empty interior. Even simpler, consider $\{0\}$. $\endgroup$ – Bungo Sep 25 '18 at 1:07
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    $\begingroup$ Try to execute your proof in the case $F=[0,1] \cup \{2\}$. You should see it go wrong in at least 3 places. $\endgroup$ – Randall Sep 25 '18 at 1:07
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    $\begingroup$ @Zduff It is not. :-) $\endgroup$ – btilly Sep 25 '18 at 1:10
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    $\begingroup$ No, the Cantor set is uncountable. $\endgroup$ – Bungo Sep 25 '18 at 1:10
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Here is a pointer to a valid proof.

The case of an empty set is trivial.

For a non-empty set, you can map each rational number onto the nearest element in F, breaking ties in favor of the smaller number.

Prove that the set of points E you have mapped the rationals onto is countable, is a subset of F, and that its closure includes every point in F.

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  • $\begingroup$ So how about: for any isolated point of $F$ let $r \in R$ be the next largest rational number. Since $R \subset \mathbb{Q}$, $R$ is countable. Now, let $\hat F$ be the interior of the union of all intervals in $F$. $\hat F$ is countable since it's open $\implies \hat F \cup R = E$ is countable. $\bar E = E \cup \{\text{limit points of } E\} \implies \bar E = F$ $\endgroup$ – Zduff Sep 25 '18 at 1:33
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    $\begingroup$ @Zduff There is no "next largest rational number". $\endgroup$ – Bungo Sep 25 '18 at 1:44
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    $\begingroup$ @Zduff There is no "next largest rational number". But what there is is an enumeration of rational numbers. So it is well defined to talk about the first rational number that meets a given condition. That fact will be useful to generate a sequence $x_i$ from $E$ that converges to a given $x$ in $F$. And with that I'll stop spoonfeeding you a potential answer to your homework problem. $\endgroup$ – btilly Sep 25 '18 at 2:44
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The comments have discussed several holes in your proof. Here is a hint for a proof to the general case, although you will have to make some extensions. Suppose $F$ was bounded, say $F\subset [a,b]$. We succesively take finite sets $E_i$, and at the end will put $E=\cup E_i$. Stage $0$ will be $E_0=\emptyset$. Once we have created $E_i$, we partition the interval $[a,b]$ into $i+1$ subintervals, $[a,a+\frac{b-a}{i+1}],[a+\frac{b-a}{i+1},a+2\frac{b-a}{i+1}],...,[a+i\frac{b-a}{i+1},b]$, and from each such subinterval we select one single point, $p_k\in F$, if $F$ intersects nontrivially with that subinterval. We then set $$ E_{i+1}=E_i\cup \{p_1,\dots,p_{i+1}\}. $$ Can you show that in this case $E=\cup_i E_i$ is countable and satisfies $\overline{E}=F$?

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  • $\begingroup$ This works, but I like avoiding the use of the axiom of choice when it is not needed. Which is why I'd use a well-defined choice function rather than saying "for each subinterval we select one single point". $\endgroup$ – btilly Sep 25 '18 at 2:45
  • $\begingroup$ @btilly Sure, you could easily adapt the above proof to avoid choice, I just don't care too much about it $\endgroup$ – TomGrubb Sep 25 '18 at 3:36

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