Every Closed Set in R has a Countable Dense Subset

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In summary, we want to show that every closed set in $\mathbb{R}$ has a countable dense subset. We can consider the case when the set is bounded, in which case it is compact and we can use finite $1/n$-nets to construct a countable dense subset. For the general case, we can write the set as a union of closed sets on bounded intervals and use the same method to show that each subset is separable, hence the entire set is separable. This gives us a countable dense subset for the unbounded set as well.
  • #1
joypav
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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!
 
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  • #2
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)?
 
  • #3
Euge said:
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.
 
  • #4
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.
 
  • #5
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?
 
  • #6
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 $E$.
 
  • #7
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?
 
  • #8
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.
 

FAQ: Every Closed Set in R has a Countable Dense Subset

1. What is a closed set in R?

A closed set in R (the set of real numbers) is a set that contains all of its limit points. This means that for any sequence of numbers in the set that converges, the limit of that sequence is also in the set.

2. What does it mean for a set to be countable?

A countable set is a set that can be put into a one-to-one correspondence with the set of natural numbers (1, 2, 3, ...). In other words, the elements of the set can be counted and listed in a specific order.

3. What is a dense subset?

A dense subset of a set is a subset that contains points that are arbitrarily close to every point in the original set. In other words, for any point in the original set, there exists a point in the dense subset that is as close to it as desired.

4. Why is it important for a closed set in R to have a countable dense subset?

Having a countable dense subset in a closed set is important because it allows for a more complete understanding of the set. It helps to fill in any "gaps" in the set and allows for a more precise description of the set's properties.

5. Can every closed set in R have an uncountable dense subset?

No, every closed set in R cannot have an uncountable dense subset. This is because the set of real numbers is uncountable, so if a set has an uncountable dense subset, it would contain more elements than the entire set of real numbers, which is not possible.

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