All right /sci/entists? Who here is able to do basic set theory? And who is just here to post memes?
Prove that if X has a smaller (possibly infinite) cardinality than Y, then [math]2^X[/math] has a smaller cardinality than [math]2^Y[/math].
Do your own homework.
>>7691670
Just create a bijection between the smaller powerset and some strict subset of the larger powerset.
>>7691685
No because an infinite set can be bijective to a strict subset of itself.
>>7691670
If both are finite, just count the elements.
If one is finite and the other infinite, trivial.
If both are infinite, statement is wrong because infinity = infinity.
>>7692139
There's more than one size of infinity.
A set is not bijective with its powerset, the powerset has a strictly greater cardinality. For example if X = [math]{\bf N}[/math], Y = [math]2^X[/math], then X has a smaller cardinality that Y and [math]2^X[/math] has a smaller cardinality than [math]2^Y[/math].
If smaller means smaller or equal then the proof is trivial. If smaller means strictly smaller, then it is independent from ZFC.
Finite case: if 2^card(X) < 2^card(Y) then card(X) < card(Y) by definition.
Infinite case: I don't know.
>>7692139
>If both are infinite, statement is wrong because infinity = infinity.
But that's wrong Anon, and you should feel bad about yourself.
>>7692358
Sorry I misunderstood OP's question, this proof is wrong. Disregard it.
>>7691670
>he uses 2^X instead of P(X)
Assuming the generalized continuum hypothesis...
>>7692403
>he uses P(x) instead of [math]\mathbb{P}(x)[/math]
>>7692483
> P(x)
> Not [math]{\cal P}(x)[/math]
>>7692647
> [math] \mathcal{P}(X) [/math]
>not [math] \wp(X) [/math]
>>7692483
But then how would I denote projective space? Let's be honest, one of these things is way more important than the other.
Ok, we have an injective function from x into y, so let S be a subset of X, we know that each element of X maps to an element of Y, so we know that this uniquely defines a subset of Y, hence we have an injective function from P(X) into P(Y).
Math major btw.
>>7692672
Oh, strict, nm. That is harder.
>>7692706
Its actually impossible to answer, rofl.
>>7692403
Fuck, beat me to it. Came here to post exactly this.
You can't prove that: Assuming MA + ¬CH, [math]2^{\aleph_0} = 2^{\aleph_1} = \mathfrak{c}[/math]
I saw it once in an book on Analysis, so it must be true. QEDeferred.
