# Can two uncountable disjoint sets be dense in $[0,1]$?

I am not sure whether this question was already asked? Let me know if so.

Suppose $$A_1$$ and $$A_2$$ are uncountable disjoint subsets of $$A$$. Can $$A_1$$ and $$A_2$$ be dense (meaning "closely approximate all points") in $$[0,1]$$? Is it possible to give an elementary example? I haven't studied real analysis in college yet.

Edit ——————

$$\require{enclose} \enclose{horizontalstrike}{A_1=\lim_{n\to\infty}\bigcup_{i=1}^{ \lceil n/2 \rceil}[0,2i/n]}$$

$$\require{enclose} \enclose{horizontalstrike}{A_2=\lim_{n\to\infty}\bigcup_{i=1}^{\lceil n/2 \rceil}[2i/n,(2i+1)/n]}$$

Would $$A_1$$ and $$A_2$$ be uncountable?

## Second Edit:

Here's what I really meant

$$A_1=\lim_{n\to\infty}\bigcup_{i=1}^{n}[(2i-2)/2n,(2i-1)/2n)]$$

$$A_2=\lim_{n\to\infty}\bigcup_{i=1}^{n}[(2i-1)/2n,2i/2n]$$

Are $$A_1$$ and $$A_2$$ uncountable?

• What are $f_1$ and $f_2$? How is the function $f$ related to the question that you are asking? Commented May 3, 2020 at 13:53
• @JoséCarlosSantos Made edits. Commented May 3, 2020 at 13:55
• The question you linked shows that in fact one can have uncountably many disjoint uncountable dense sets, even with the additional requirement that the sets be measurable. I don't consider your question a duplicate, though, since you ask for an elementary example of two such sets. I don't have such an example for you, though. Commented May 3, 2020 at 14:07

Let

$$B_1 = \left (0, \frac{1}{2} \right) \cap \mathbb Q$$

$$B_2 = \left( 0, \frac{1}{2} \right) - \mathbb Q$$

$$B_3 = \left(\frac{1}{2}, 1 \right) \cap \mathbb Q$$

$$B_4 = \left(\frac{1}{2}, 1 \right) - \mathbb Q$$

Then

$$A_1 = B_1 \cup B_4$$

$$A_2 = B_2 \cup B_3$$

• I'm accepting your answer. See my answer below. Is mine also correct? Commented May 3, 2020 at 14:28

Yes. In fact, you can find an uncountable collection of uncountable dense sets.

The article “Partitioning the Real Line into an Uncountable Collection of Everywhere Uncountably Dense Sets” by Seth Zimmerman and Chungwu Ho in The American Mathematical Monthly (vol. 126, no. 9, November 2019, p.825) will be of interest.

• Thank You for the article. Is my answer below correct? Commented May 3, 2020 at 14:30
• Wow, I was pretty shocked to see a published article about such a trivial problem. I found the abstract quite hilarious: "A recent paper showed that..." and then something that could have been an exercise in the early 19 hundreds. But maybe this is more common than I think... Commented May 3, 2020 at 15:17

$$\require{enclose} \enclose{horizontalstrike}{A_1=\lim_{n\to\infty}\bigcup_{i=1}^{ \lceil n/2 \rceil}[0,2i/n]}$$

$$\require{enclose} \enclose{horizontalstrike}{A_2=\lim_{n\to\infty}\bigcup_{i=1}^{\lceil n/2 \rceil}[2i/n,(2i+1)/n]}$$

Would $$A_1$$ and $$A_2$$ be uncountable?

What I meant was

We can generalize the process as

$$A_1=\lim_{n\to\infty}\bigcup_{i=1}^{n}[(2i-2)/2n,(2i-1)/2n)]$$

$$A_2=\lim_{n\to\infty}\bigcup_{i=1}^{n}[(2i-1)/2n,2i/2n]$$

• Note that your $A_=[0,1/2]$, so there’s already a problem...
– MPW
Commented May 3, 2020 at 14:55
• Don’t ask a question in an answer to your own question Commented May 3, 2020 at 15:29
• @MPW Made corrections Commented May 3, 2020 at 15:59
• Not sure what you're trying for, but of course $\bigcup_{i=1}^{n}[0,a_i] = [0,a_n]$ if $a_i$ is an increasing sequence. This "fix" didn't fix anything.
– MPW
Commented May 3, 2020 at 16:03
• @MPW $[0,a_i]$ is an interval. Now suppose in the first “step” $A_1=[0,1/2]$ and $A_2=[1/2,1]$. In the second “step” $A_1=[0,1/4]\cup[1/2,3/4]$ and $A_2=[1/4,1/2]\cup[3/4,1]$. In the third “step”, $A_1=[0,1/8]\cup[2/8,3/8]\cup[4/8,5/8]\cup[6/8,7/8]$ and $A_2=[1/8,2/8]\cup[3/8,4/8]\cup[5/8,6/8]\cup[7/8,1]$. As the iterations approach infinity we get our final $A_1$ and $A_2$. Are $A_1$ and $A_2$ uncountable? Commented May 3, 2020 at 16:28