Is the axiom of choice a cheat?

>>8902140
Define cheat

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>mfw people think it makes sense for uncountably infinite dimensional vector spaces to have a basis

>>8902146
Brainlet confirmed
>>8902140
Yes. Every set contains the void element.

>>8902140
It kinda is. The axiom of choice (and the theorems it implies) are a way to apply finite and countable logic, to uncountable sets.

A good example is the well-ordering principle. A proposition perfectly equivalent to the axiom of choice. It simply says that every set can be well ordered.

If I gave you a finite set, finding such an order would be trivial.

If I gave you a countably infinite set, finding such an order would be a neat exercise but still trivial. Maybe you would start by defining a bijection from that set to the natural numbers and then we all know what a good well ordering of natural numbers is.

But the real numbers? Impossible to construct. To this day no one can give a well ordering for the real numbers nor for any other uncountable set.

BUT WOULDN'T IT BE NICE IF WE COULD?

And when I say that it would be nice, I don't mean that it would be nice to be able to construct it. I mean that here I have a little theorem of mine. This little theorem is about sets and all it needs is that the set can be well ordered. This little theorem is really fun and wouldn't it be nice if this little theorem was true for all sets? Even if they are uncountable? WOULDN'T THAT BE FUN?

>>8902140
I am no mathematician, so disregard my opinion, but concretely it's shit, but practically it is indispensable. I'm not sure if people have been able to construct a set where it is impossible to choose an element of that set, but something tells me the number of those kinds of sets faaaaar outstrip nicely behaved sets, it's just hard to say if mathematics is useful in those settings. For all practical purposes though, it's a shortcut to being rigorous about something that almost always you can take for granted.

>>8902140
Did you choose to ask this or did you decide before you became aware of the fact that you did?

>>8902154
Even that one set containg all the sets not containing itself?
Or we Zermelo Fraekel now?

>>8902351
Yes, aand that is by definition of void element.

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>>8902146
go ahead and explain how it doesn't make sense

>>8902154
>>8902548
>void element
what the fuck is this pseudotry?

>Is the axiom of choice a cheat?
yes

It's a make-pretend axiom to make functional analysis look more like a theory of finite sets.

The axiom of choice is right and true for finite sets, so in that sense I'm very much pro-AoC.
However, many set theories like ZFC don't capture the notion of "set" well. They descibe something that has some properties of what we understand as sets and augments it with more stuff. Superlarge sets are an example. For these, the AoC just induces some nice looking statements, but they have no practical meaning. They make the theory just look more familiar and with less holes (theorems we can't decide).
In short, the "problem" (if you will) is to a the power-set axiom to your set theory, so that the chain on infinite sets
A=N
B=PN
C=PPN
D=PPPN
E=PPPPN
F=...
is one where each set has a larger cardinality than ther previous.
The set of functions from PN to {0,1}, i.e. something like 2^R, is already something we can't really comprehend as something of size.

The argument
>but AoC is equivalent to
>the Cartesian product of non-empty sets is non-empty
is polemical or misguided.
Here, people confuse "it's not provable that there exists a x so that P(x)" with "there is no x such that P(x)".
With choice you can prove that the product has an element.
Without choice, you can't prove that it has an element. This doesn't mean that there are non-empty sets whos product are empty.

>>8902581
New math, I just made it up. My Iq is very high

>>8902328
>a set where it is impossible to choose an element of that set

don't answer the question if you don't know what you're talking about

>>8902609
>It's a make-pretend axiom
just like every other axiom. thanks for making a completely useless post

>>8902667
god i hate you

AOC is equivalent to the statement "A non-empty product of non-empty sets is non-empty." pretty fucking obvious to me.

>>8902181
>To this day no one can give a well ordering for the real numbers nor for any other uncountable set.
Wrong. The ordinal are very easy to well order And you don't even need AC.

>>8902773
The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?

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>>8902146
>mfw C[a,b] for all intents and purposes has no basis

>>8902864
>well-ordering principle obviously false
Support that claim. It's more intuitive than the Axiom of Choice to me.

>>8902841
The ordinals are fucking trivial man.

Well order me my reals and we will be talking. I'll make it easier for you, give me a well ordering of (0,1)

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>>8902181
>To this day no one can give a well ordering for the real numbers

>>8903173
Well, my friend. Give me a well ordering of the real numbers. I have connections and I can get you that fields medal mailed in less than 2 weeks. Just make the post.

>>8903173
There's a literal proof you cannot construct a well ordering on the reals with the language of set theory.

>>8903168
Yeah, but you just claimed no one can construct a well ordering of any other uncountable set. This is plainly false.

>>8903185
no reals share an equivalence class. rest of the proof is trivial

ill be waiting for that medal

>>8903201
The rest of the proof? What proof?
I asked for a construction and you gave me a half asses "argument".

I guess it is Terry Tao again this year.

>>8903203
It is impossible to construct, yet it exists.

>>8903185
Why don't you say explicitly what kind of description of the well-ordering would qualify.

>>8903168
>>8903185
take your favorite bijection from R to the first uncountable ordinal and there you go

or what? me using the same trick you used with the axiom of infinity is outlawed because it hurts your feels?

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>>8903260
>assuming the continuum hypothesis

>>8903279
take the appropriate ordinal then you autist, whatever you believe it to be

>>8903260
Teaching set theory to freshmen was a mistake

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>there are people on this planet RIGHT NOW that fell for cantor's diagonal arguement

>>8903294
Give a bijection then.

>>8903295
giving that to a brainlet such as yourself would be casting pearls before swine, but i will give you something to activate your almonds

consider a bijection between N and N:

1 = (8)3463494362345234.....
2 = 5(7)856834834836836.....
3 = 43(3)52567234234215.....
n = ....
....

construct the number k, where k is different from n in the n'th digit, e.g. 521...

k is an element of N and is not an element of this bijection, therefor, there is no bijection between N and N

>>8903316
your picture is misleading, natural numbers don't have infinite digits
whatever "number" you build would need to be infinitely long

>>8903287
>If you fix an ordinal α, then it is consistent that c>ℵα. More precisely, there is a (forcing) extension of the universe of sets with the same cardinals where the inequality holds.

https://math.stackexchange.com/questions/20385/bound-on-the-cardinality-of-the-continuum-i-hope-not

makes u think

>>8903342
depends on your construction of N

you could have every natural have infinite digits using leading zeroes

>>8903374
I clearly mean significant digits which is well defined
and it's obvious that your "number" k is not a natural, no matter what your "construction of N" is

>>8903356
as interesting and mindblowing as this is, if you already fixed a universe were R exists then choice implies a bijection between it and some ordinal

>>8903381

even without infinite digits, k would be an element of N and not an element of the bijection

>>8903397
are you pretending not to understand me on purpose? I can't be any clearer. your "number" k is not a natural. it has infinite "wrong" digits, no matter how you understand that or what exotic construction of N you want to use

>>8903400
i know you're a literal brainlet but i just said you could do the same with finite digits

1 = (5)32
2 = 5(3)15
3 = 66(4)62
4 = 437(2)37
...

k = 2153... (finite or infinite, end wherever you want to)

k will be an element of N and not within any bijection

>>8903410
I can also say that 1 + 1 = 3, that doesn't make it true
honestly just fuck off, you're too dumb and lazy to be able to look at what you're fucking saying critically

>>8903434
you don't feel any shame or embarrassment from posting something like that?

>>8903316
This is unrigorous freshman tier reasoning. Let me show you why

Your k is not well defined. Note that if your list has all the natural numbers, and k has to be different from all of those, how many digits will it have? Infinitely many! Thus k is not a natural number. This is a really simple mistake even a brainlet could spot so I fear for you.

>>8903607
you can't even necessarily construct any such k, unless the ith natural number on the list always has at least i digits, which is dumb

>>8903617
True

>>8903607
>>8903617

i know you're literal brainlets but i just said you can do the same with finite digits. i will try this one last time at a level around elementary.

1 = 1
2 = 11
3 = 111
n = 10^n - 1 / 9

k = 2

n is finite by definition of naturals

if n is finite and natural, 10^n - 1 / 9 is finite and natural by definitions of arithmetic

every imput is a natural number
every output is a natural number
2 is a natural number

N is not mapped to k in this bijection

there is no bijection between N and N