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May 2nd, 2019, 05:40 PM   #1
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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).
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May 2nd, 2019, 06:12 PM   #2
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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.
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May 2nd, 2019, 06:42 PM   #3
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Tax season's over I guess.

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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.
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May 2nd, 2019, 06:51 PM   #4
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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.
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May 2nd, 2019, 06:55 PM   #5
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Quote:
Originally Posted by AplanisTophet View Post
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.
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May 2nd, 2019, 07:02 PM   #6
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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.
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May 2nd, 2019, 07:05 PM   #7
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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.
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May 2nd, 2019, 07:21 PM   #8
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Quote:
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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 $V \cap \mathcal{C} = V$?

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Would you say that the intersection of a set with any other set is a subset of each set respectively?
Wrong thread.

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Originally Posted by Maschke View Post
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.

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I'm disappointed I had to explain this to you.
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May 2nd, 2019, 07:29 PM   #9
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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.
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May 2nd, 2019, 07:29 PM   #10
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Quote:
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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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