A taste of abstract mathematics - Vitali set

[@Anonymous_MC]

kinda crazy to see how much this channel has changed...

[@yamunan-bitset9388]

Nice video! You should make a video on the Dedekind Cut.

[@dedekindcuts3589]

I have been thinking about it!!

[@nikitasmarkantes5046]

Thank you so much. A really hidden gem..

[@Smoothcurveup52]

Wonderful explaination

[@nestorwalters8079]

Great video, thank you!

[@suhaibalkhaldi]

Thank you so much

[@youngone999]

설명이 명확하고 직관적이네요, 생각이 잘 정리됩니다, 감사합니다, 다만 아래에 대한 설명이 없네요 [0,1]⊂ ..

[@gaaraofddarkness]

3:23 are the different colored sets disjoint?

[@dedekindcuts3589]

Yes, each colored set is supposed to represent an equivalence class. For the rule that describes how we partition [0,1] into disjoint equivalence classes, the rule is that a ~ b if and only if a-b is rational. One can check that this rule is reflexive, symmetric and transitive.

[@Anonymous-df8it]

Couldn't you use infinitesimals to define the length of the Vitali set? So, the length of said set would be between epsilon and 3*epsilon.

[@rationalsolver9771]

The idea is to define a length function with codomain a real number, and what you mean by infinitesimals are not real numbers

[@Anonymous-df8it]

@@rationalsolver9771 What's 'codomain'?

[@rationalsolver9771]

@@Anonymous-df8it the values which the function can return, when one writes: Let the function f: A -> B, the codomain is B

[@MikeRosoftJH]

As another user has said, the measure of a set (under the standard definition of measure) is a non-negative real number, or infinity (meaning that the set can't be covered by any finite or countably infinite collection of intervals of a finite total length). Intuitively speaking, the measure of the Vitali set is "infinitely small"; but does it help to extend the definition of measure to include infinitesimals? I'd say: not very much, because this still wouldn't be countably additive. Again, we have shown that the union of countably many shifted copies of the Vitali set is a superset of an interval of length 1, but a subset of an interval of length 3. Now consider this: let's put the sets in a sequence (we can do this, because the collection is countably infinite), and observe: the first set has measure less than 1/4 (because it's infinitesimal), the measure of the second is less than 1/8, the third is less than 1/16, the fourth less than 1/32, and so on. So the total is less than 1/4 + 1/8 + 1/16 + 1/32 + ...; that's a geometric series with sum 1/2 (the value of an infinite sum is a limit of the sequence of partial sums). So the sum of the measures is less than 1/2 (indeed it's less than any positive real number, by picking an arbitrarily small upper bound of the first measure); yet the union is a superset of an interval of length 1, and a subset of an interval of length 3 (so the measure of the union - if the union is measurable - should be between 1 and 3). I think that it would have been more useful to say that the measure of the Vitali set is 0, under some finitely additive measure which extends the usual (countably additive) Lebesgue measure.

[@hacker2ish]

Interesting, i wonder if there's an immeasurable set that can constructed without the axiom of choice. Maybe all such sets must be measurable?

[@MikeRosoftJH]

In absence of axiom of choice, it's consistent that all sets of reals are measurable (assuming that set theory itself is consistent). And even though under axiom of choice (or under assumption that continuum can be well-ordered) a non-measurable set is guaranteed to exist, no formula can be proven to define one (that is, there is no formula for which the following can be proven: 1) there exists exactly one set which satisfies the formula, and 2) that set is a non-measurable set of reals).

[@karimkafi1872]

Good explanation, just one thing if only you have mentioned the axiom of choice in the construction of the set V.

[@dedekindcuts3589]

Ah yes, it uses the axiom of choice!

[@MikeRosoftJH]

@@dedekindcuts3589 It follows that if the choice set doesn't exist, then there are more equivalence classes than points. (Weird, isn't it? That no set can be partitioned into more disjoint, non-empty subsets than it has elements is a consequence of the axiom of choice.)

[@Anonymous-df8it]

He did. Just not by name