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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 ——————

What about...

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

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

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?

Arbuja
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    What are $f_1$ and $f_2$? How is the function $f$ related to the question that you are asking? – José Carlos Santos May 03 '20 at 13:53
  • @JoséCarlosSantos Made edits. – Arbuja May 03 '20 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. – saulspatz May 03 '20 at 14:07

3 Answers3

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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 $

$A_2 = B_2 \cup B_3$

Coriolanus
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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.

MPW
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  • Thank You for the article. Is my answer below correct? – Arbuja May 03 '20 at 14:30
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    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... – Jonathan Schilhan May 03 '20 at 15:17
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What about...

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

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

Would $A_1$ and $A_2$ be uncountable?

What I meant was

We can generalize the process as

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

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

Arbuja
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    Note that your $A_=[0,1/2]$, so there’s already a problem... – MPW May 03 '20 at 14:55
  • Don’t ask a question in an answer to your own question – gen-ℤ ready to perish May 03 '20 at 15:29
  • @MPW Made corrections – Arbuja May 03 '20 at 15:59
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    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 May 03 '20 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? – Arbuja May 03 '20 at 16:28