Is this a meme?

Keep telling yourself that senpai. ;)

>everyone who ever touched higher level math must be comfortable with an existence of a bijection between integers and even integers - this means that we can split Z and get two copies of Z. banach-tarski is actually just a very delicate usage of the same principle.
No you are just a retard who watch youtube videos and think he's an expert, banach-traski has nothing to do with bijections, it uses isometries wich are WAY stricter transformations, and there is no isometry between the integers and even integers.
In fact there is no duplicable set in R.

>wow look at me i have no critical thinking skills but think im smarter than everyone else
lol

Libshit major.

nah, just a humble mathfag
enjoy being a pseud and never accomplishing anything in life

But he's right though. The integers are countable. A partition of Z into any number of subsets which are each isomorphic to Z is trivial, both because Z is countable and because these bijections are isomorphisms which preserve structure, and Z doesn't have the same notion of measure that Euclidean space does. Banach-Tarski IS a very surprising result because it shows you can decompose a ball into finitely many pieces and compose these into two balls of the same measure using only isometries, meaning that distance and measure are invariant throughout this entire process. This is indeed very strange and is a classic example of the strangeness permitted by the axiom of choice.

You call yourself a humble mathfag but if you really want to be a mathematician the first lesson to learn is that you will be wrong an overwhelming proportion of the time when talking about things near the boundary of your knowledge. That's okay, but if you value winning an argument more than being correct and developing a useful understanding and intuition of a problem or concept, then I'm afraid it's YOU who are the pseudointellectual brainlet, and you will never accomplish anything in mathematics beyond the undergraduate level. Apply yourself, drop out, or switch back to electrical engineering. We have no room for you here.

>is actually just a very delicate usage of the same principle
>can't understand drawing comparisons between two similar scenarios
please keep making an embarrassment of yourself

Why is this "principle" not possible in dim 1 then ? or dim 2 with a bounded set ?

Not enough space to move around.

Nope, you can't duplicate a segment even in R^3 with as much "space to move around" than with the paradoxe.

A line in R^3 is not making proper use of the available extra space.

>implying a statement gets stronger with requirements

I never got why that sphere duplication was such a mysterious thing.

It's easy to show that kind of thing on a small, understandable scale.

For example, you can take the integers and partition them into two disjoint sets: the set of even integers, and the set of odd integers. Now take every element in the even set and divide it by two; and then take every element in the odd set and subtract one and divide it by 2. Now look what you've done -- you started with the set of integers and you've created two duplicate copies of it.

That ability to duplicate an infinite set is really just a natural consequence of the way you prove that two infinite sets have the same cardinality (by showing that a bijective function exists between the two sets).

If I can do this easily with the set of integers, then why should I be so surprised that you can also do a similar thing (albeit more complicated) with the set of points on a sphere?

It has to do with the fact that the reassembly is a bunch of isometries.

All you're allowed to do is translate and rotate things. In your integer example, you divided by 2. That's definitely not an isometry of Euclidean space as it stretches/shrinks things. Imagine trying to create two copies of a rigid ball by cutting it into pieces and fitting them back together. No stretching.

>The Axiom of Choice is obviously true; the Well Ordering principle is obviously false; and who can tell about Zorn’s lemma?

Haha... I get it... so funny.... :/

Anyway which one of your nerds can help BanachTarski my bank account?

Invest everything in bitcoin, cash out in 1 year.
You're welcome.

To elaborate, no one is surprised that there's a bijection between the single sphere and the two copies of the sphere. As you've said, that follows naturally from the two sets having the same cardinality.

Its just the whole "we think isometries should preserve areas and volumes" thing.

And since I have to run and probably won't be back here for a few weeks, I'll end with this.

The real paradox of BT is that taking the Axiom of Choice leads to the existence of non-measurable sets (not sets with measure zero like points, but sets without well-defined measure) which allow for volumes to change when objects are translated and rotated.

And that's pretty fucked.

However, since I don't really care about nonmeasurable sets, I take AC anyway and go work elsewhere. Down with analysis, up with arithmetic geometry!

he's right, you know

>necessary existence of non-measurable sets.
necessity is a spook for undergraduates

...

is this a meme?

It's a cool meme.

it's mathematic's schroedinger's cat. it's an intuitively absurd but valid application of the underlying theory. it should show any sane person that the underlying theory is flawed

>tfw remember when bitcoins were hovering around $5 a coin and thought they were going to die in six months
>tfw they spiked to $15 one time and i thought there might be something there but didn't have any money lying around to blow

Brainlet version math.hmc.edu/funfacts/ffiles/30001.1-2-8.shtml

only it's not a metaphore you doofus and you cannot do this in real life
show me a non measurable set and we'll talk. the theory is good

This is by far the best one I've seen of these.

>and you cannot do this in real life
That's the point.

>All you're allowed to do is translate and rotate things. In your integer example, you divided by 2. That's definitely not an isometry of Euclidean space as it stretches/shrinks things. Imagine trying to create two copies of a rigid ball by cutting it into pieces and fitting them back together. No stretching.

That kind of loses its weight once you realize that at least one "piece" must consist of infinitely many cuts that are made in a manner that cannot be unambiguously described using any formula of finite length.

Can someone explain the axiom of choice in a brainlet-friendly manner?

It simply states that for any set (called X) of non-empty sets, there is a function that matches any set within X an object that belongs to it.

You'd already have won 10% of your invested summ already if you followed this advice at that time btw.

What said is right, but to expand a little.

If I think about the collection of closed intervals [a,b] on the real line, there is a function which maps each interval to a point in the interval. Namely f([a,b]) = (a+b)/2. The function f here is called a "choice function". It's a function we takes our collection of sets (in this case the collection of closed intervals), and maps each set to an element of the set.

More generally, if X is a collection of non-empty sets, a choice function is a function [math] f: X \to \mathcal{P}(X)[/math], with the property that for all [math] A \in X[/math] we have [math]f(A) \in A[/math].

In the example above, we explicitly constructed a choice function, we didn't need a special axiom to tell us a choice function existed. There are other cases in which no choice function can be constructed explicitly. For example, put [math]X = \mathcal{P}(\mathbb{R}) \setminus \{\emptyset\}[/math]. Try and think about how you could define a function which maps every subset of [math]\mathbb{R}[/math] to an element of itself. You specifically would need a "rule" for your definition, like we did in the case of the interval [a,b]. You should be able to convince yourself that this is impossible.

In a case like this, we could invoke axiom of choice and assert a choice function exists. It may seem intuitive that such a function ought to always exist, but besides the fact that axiom of choice is equivalent to less intuitive principles like well ordering principle, you also have the problem that invoking this axiom is non-constructive. For example, in Banach-Tarski in particular, axiom of choice is used in a way which is unavoidable, so while you can prove a decomposition of a sphere into two sphere exists formally, it is not possible to actually construct those sets, i.e., write explicit definitions such that you can say which points are and are not in the sets of decomposition.

hmm i think there is alot of non measurable sets.
i you cant see them then dont go for that theory

Nice meme

very good post, thanks

Scientists of wutology envision this as how the big bang found its start

Infiniti divided by two equals infiniti