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May 2nd, 2019, 05:40 PM  #1 
Senior Member Joined: Jun 2014 From: USA Posts: 528 Thanks: 43  Intersection of Two Sets Question
Does there exist a Vitali set $V$ such that the difference between $V$ and the Cantor ternary set is empty? Vitali set $V$: https://en.wikipedia.org/wiki/Vitali_set Cantor ternary set $\mathcal{C}$: https://en.wikipedia.org/wiki/Cantor_set Equivalently, is the following statement true? $$\exists V \in \{y : y \text{ is a Vitali set}\}(x \in V \implies x \in \mathcal{C})$$ I don't believe there is a Vitali set that is a subset of $\mathcal{C}$ because the Lebesgue Measure of $\mathcal{C}$ is $0$ while the measure of a Vitali set is undefined (if not positive in the sense that the union of countably many Vitali sets may have a positive real measure so it's possible to consider the measure of a single Vitali set as being both positive and infinitesimal in addition to being nonLebesgue measurable). 
May 2nd, 2019, 06:12 PM  #2 
Senior Member Joined: Jun 2014 From: USA Posts: 528 Thanks: 43 
I'll go as far as to conjecture that the difference between any Vitali set and the Cantor ternary set will always be itself an uncountable set.

May 2nd, 2019, 06:42 PM  #3 
Senior Member Joined: Aug 2012 Posts: 2,357 Thanks: 740 
Tax season's over I guess. Bad logic. The measure of a Vitali set intersected with the rationals is zero. Last edited by Maschke; May 2nd, 2019 at 06:47 PM. 
May 2nd, 2019, 06:51 PM  #4 
Senior Member Joined: Jun 2014 From: USA Posts: 528 Thanks: 43  Bad logic, because $\mathcal{C}$ is uncountable and contains all numbers in [0,1] that may be expressed using only 0's and 2's in ternary notation. In other words, I'm not intersecting a Vitali set with the rationals... that intersection would obviously have a cardinality of 1 silly goose.

May 2nd, 2019, 06:55 PM  #5  
Senior Member Joined: Aug 2012 Posts: 2,357 Thanks: 740  Quote:
 
May 2nd, 2019, 07:02 PM  #6 
Senior Member Joined: Jun 2014 From: USA Posts: 528 Thanks: 43  I have, which is why I came here, seeing if someone else would too. You haven't. Read the OP again if you like, but you're posting in the wrong thread or something it seems because what you wrote was in no way applicable.

May 2nd, 2019, 07:05 PM  #7  
Senior Member Joined: Aug 2012 Posts: 2,357 Thanks: 740  Quote:
And what would be the measure of any subset of a set of measure zero? I'm disappointed I had to explain this to you. Last edited by Maschke; May 2nd, 2019 at 07:10 PM.  
May 2nd, 2019, 07:21 PM  #8  
Senior Member Joined: Jun 2014 From: USA Posts: 528 Thanks: 43  Quote:
Does there exist a Vitali set $V$ such that $V \cap \mathcal{C} = V$? Quote:
Quote:
 
May 2nd, 2019, 07:29 PM  #9 
Senior Member Joined: Jun 2014 From: USA Posts: 528 Thanks: 43  I get it now, you were neglecting the possibility that the measure of the intersection between $V$ and $\mathcal{C}$ could be undefined, especially if the intersection were to equal $V$. You can't just assume that the intersection would have a measure of $0$ simply because $\mathcal{C}$ does as this says nothing of whether or not the measure of the intersection may be undefined.

May 2nd, 2019, 07:29 PM  #10 
Senior Member Joined: Aug 2012 Posts: 2,357 Thanks: 740  Didn't you say it was 1? And throw in a gratuitous mild personal insult? I'm sure you know I can sling shit if that's what somebody wants.


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