Every Closed Set in R has a Countable Dense Subset

[joypav]

Show that every closed set in R has a countable dense subset.
Let's call the set F.

I've been thinking about this problem for a little bit, and it just doesn't seem like I have enough initial information!
I tried listing some things that I know about closed sets in R:

$\cdot$ Countable dense subset is the same as being separable (I think?)
$\cdot$ F contains all of it's limit points
$\cdot$ Every cauchy sequence in F converges to a point in F
$\cdot$ If F is closed then F is $G_\delta$, a countable intersection of open sets (I thought this may be helpful because the sets are countable?)

I considered somehow utilizing the rational numbers, because they are dense in R, and for each x in F having a sequence of rational numbers that converges to x. However, this was just an idea.

Obviously I haven't done much concrete work on it. But I would appreciate a push in the right direction!

 

[Euge]

Hi joypav,

First consider the case when $E$ is bounded. Then $E$ is compact. Can you prove in this case, that $E$ is separable (i.e., has a countable dense subset)?

 

[joypav]

Hi joypav,

First consider the case when $E$ is bounded. Then $E$ is compact. Can you prove in this case, that $E$ is separable (i.e., has a countable dense subset)?

If E is compact, then every open cover has a finite subcover. Then, basically, you can have a union of the center of each "ball" of your finite subcover. (by shrinking the radius) And that would be a dense and countable subset.

 

[Euge]

Not quite. Just one finite subcover will not give density. But for each $n\in \Bbb N$, we may consider finitely many centers of balls of radius $1/n$ covering $E$ (which forms a finite $1/n$-net), and let $A$ be the union of all these centers. Then we can show that $A$ is dense in $E$. Being the countable union of finite sets, $A$ is countable.

 

[joypav]

Here we go...

Let $E \subset R$, E closed.
Assume E is bounded $\implies$ E is compact $\implies$ every open cover of E has a finite subcover.
Consider the open cover,

$E \subset \cup_{x \in E}B(x,1/n)$, for $n \in \Bbb{N}$

Then there exists a finite subcover, for some $N \geq 1$,

$E \subset \cup^{N}_{i=1}B(x_i,1/n) \subset \cup_{x \in E}B(x,1/n) $

Define $C_n = $ the set of all centers of $B(x_i,1/n)$, $1\leq i \leq N = x_1 , x_2 , ... , x_N $

Create this set for every $n \in \Bbb{N}$.
Meaning,
$E \subset \cup^{N_1}_{i=1}B(x_i,1) \subset \cup_{x \in E}B(x,1) $, $C_1 = x_1, x_2, ... , x_{N_1} $
$E \subset \cup^{N_2}_{i=1}B(x_i,1/2) \subset \cup_{x \in E}B(x,1/2) $, $C_2 = x_1, x_2, ... , x_{N_2} $
... and so on for $C_3, C_4, ...$

Let $C = \cup_{n \in \Bbb{N}} C_n$.

Claim: C is our countable dense subset of E.
1. C is countable.
$\forall n \in \Bbb{N}, C_n \subset \cup^{N_n}_{i=1}B(x_i,1/n) $ and a subset of a countable set is countable
$\implies \forall n \in \Bbb{N}, C_n $ is countable.
$C = \cup_{n \in \Bbb{N}} C_n$, a countable union of countable sets $\implies$ C is countable.

2. C is dense in E.
Take $x \in E$. Given $\epsilon > 0$, $\exists n \in \Bbb{N}$ such that $1/n < \epsilon$.
Then, $\exists y \in C_n \subset C$ so that $x \in B(y, 1/n) \implies y \in B(x,1/n) \subset B(x,\epsilon)$
So $y \in B(x,\epsilon) \cap C \implies $ C is dense in E.

How's that look?

 

[Euge]

It looks better, but it could be improved by writing the following:

If $E$ is bounded, then $E$ is compact. Thus, for every $n\in \Bbb N$, there is a finite $1/n$-net $C_n$. If $C = \bigcup C_n$, then $C$ is countable, being the countable union of finite sets. Further, if $\epsilon > 0$ and $x\in E$, we may choose an $n\in \Bbb N$ such that $1/n < \epsilon$, and select $y\in C_n$ such that $d(x,y) < 1/n$. Then $d(x,y) < \epsilon$, showing that $x\in \bar{C}$. Hence, $C$ is dense in $E$.

 

[joypav]

Then if we want to show the same for an unbounded close set, it just seems like I don't have enough info to use?

 

[Euge]

There is enough info. In the general case, write $E$ as the union of sets $E_n := E\cap [-n,n]$ for $n = 1,2,3,\ldots$, and note each $E_n$, being a closed subspace of the compact interval $[-n,n]$, is compact, hence separable. See what you can do from there.