# Construct uncountably dense and holes everywhere

1. May 8, 2013

### jostpuur

I want to define a set $X\subset [0,1]$, without using the axiom of choice, with the following property: Both $[a,b]\cap X$ and $[a,b]\backslash X$ are uncountable for all $[a,b]\subset [0,1]$ where $a<b$.

I don't know how to define $X$ so that $[a,b]\cap X$ would always be uncountable without $X$ containing some interval. But if it contains some interval, then $[a,b]\backslash X$ will be empty sometimes.

2. May 8, 2013

### jostpuur

Never mind. I invented a solution now.

Now the problem is left as a challenge to the rest.

3. May 8, 2013

### jostpuur

My solution has the property $m^*(X)=0$. I'm still interested to know if the $X$ can be defined in such way that $0<m^*(X)< 1$.

4. May 11, 2013

### jibbles

Consider $X:=\{x\in[0,1]\backslash \mathbb{Q}: \text{the decimal expansion of } x \text{ contains only finitely many, say, }0\text{'s and }9\text{'s}\}$. This set is uncountably dense in [0,1] and so is its complement, though I'm not sure about their measures.

5. May 12, 2013

### Bacle2

I think you can calculate its measure like this:

Start with 0: if there are no 0's in the expansion, then you remove from [0,1]:

i.1)The segment from 0 to 0.1 , with measure 0.1

ii.1)The segment from 0.1 to 0.11 , with measure 0.01

n.1)The segment from 0.11111 ( n 1's) to 0.111111 (n+1-ones)

i.2) The segment from 0.2 to 0.21

...........

And you do the same thing by removing the segments :

i.1') From 0.9 to 1

ii.1') From 0.09 to 0.1

etc.

6. May 13, 2013

### jibbles

I'm confused. If X is dense in [0,1], wouldn't removing segments of [0,1] also be removing chunks of X? ... or maybe i'm the one being dense.

7. May 15, 2013

### Bacle2

Actually, you are right in that I should throw the rationals back-in after removing all the
other parts. If this is what you meant, you're right --but remember that the Rationals

Modulo the above, I don't think so; you are removing chunks of [0,1] itself, and X is what remains after all these parts are removed and you throw Q back-in. After you remove all the chunks between a_n=0 and a_(n+1) for n≥1 , same for a_n=9 , then removing all the points in
the segment/interval [an,an+1 ], you will remove all the Irrationals
with either 0-or-9 in their decimal expansion, BUT you will also remove the Rationals in the segment. Then you can throw the Rationals back-in. But remember that the Rationals have
Lebesgue measure zero, so removing the whole chunk with Rationals included will not affect
the measure of the remaining set.

8. May 15, 2013

### jibbles

But there are elements of X in all those segments you're removing. Let me repeat: X is the set of all irrationals whose decimal expansion contains at most finitely many 0s and 9s. So, for example, 0.00125225222522225 ... (nonrepeating sequence) is an element of X in [0, 0.1].

9. May 15, 2013

### Bacle2

Sorry, I misread the definition of your set; I thought it was the collection of irrationals that contained neither a zero nor a nine in their expansion.