Construct uncountably dense and holes everywhere

[jostpuur]

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

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

 

[jostpuur]

Never mind. I invented a solution now.

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

 

[jostpuur]

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

 

[jibbles]

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

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

Consider [itex]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}\}[/itex]. This set is uncountably dense in [0,1] and so is its complement, though I'm not sure about their measures.

 

[Bacle2]

Consider [itex]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}\}[/itex]. This set is uncountably dense in [0,1] and so is its complement, though I'm not sure about their measures.

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.

 

[jibbles]

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.

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.

 

[Bacle2]

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.

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
have measure zero. If not, please read-on:

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 [a_n,a_n+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.

 

[jibbles]

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
have measure zero. If not, please read-on:

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 [a_n,a_n+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.

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].

 

[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.