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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 non-Lebesgue 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
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Tax season's over I guess.

Quote:
 Originally Posted by AplanisTophet 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 ...
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
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Quote:
 Originally Posted by Maschke Tax season's over I guess. Bad logic. The measure of a Vitali set intersected with the rationals is zero.
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
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Quote:
 Originally Posted by AplanisTophet 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.
Jeez Louise brother. Done here. Think about what you just wrote.

May 2nd, 2019, 07:02 PM   #6
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Quote:
 Originally Posted by Maschke Jeez Louise brother. Done here. Think about what you just wrote.
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
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Quote:
 Originally Posted by AplanisTophet 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.
Would you say that the intersection of a set with any other set is a subset of each set respectively?

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
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Quote:
 Originally Posted by Maschke Would you say that the intersection of a set with any other set is a subset of each set respectively?
I asked a very specific question relating to the intersection of two sets:

Does there exist a Vitali set $V$ such that $V \cap \mathcal{C} = V$?

Quote:
 Originally Posted by Maschke Would you say that the intersection of a set with any other set is a subset of each set respectively?

Quote:
 Originally Posted by Maschke And what would be the measure of any subset of a set of measure zero?
Zero, which is why I don't think $V$ could be a subset of $\mathcal{C}$, which is the main bullet point of this whole conversation you seem oblivious to.

Quote:
 Originally Posted by Maschke I'm disappointed I had to explain this to you.

May 2nd, 2019, 07:29 PM   #9
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Quote:
 Originally Posted by Maschke Tax season's over I guess. Bad logic. The measure of a Vitali set intersected with the rationals is zero.
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
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Quote:
 Originally Posted by AplanisTophet Zero, which is why I don't think $V$ could be a subset of $\mathcal{C}$, which is the main bullet point of this whole conversation you seem oblivious to.
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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