4 ms·
> I've never made peace with Cantor's diagonaliztion argument Maybe you'd prefer a purely set-theoretic one: --- Let R be a set. Let S be a set of all subset
by bheadmaster 1y ago
> I've never made peace with Cantor's diagonaliztion argument
Maybe you'd prefer a purely set-theoretic one:
---
Let R be a set. Let S be a set of all subsets of R.
We want to prove that |S| > |R|, by proving that a bijective function from R to S cannot exist. We will do that by assuming that it can, and then deriving a contradiction.
Assume there is a bijective function f : R -> S. Define D = { r ∈ R | r ∉ f(r) }.
Since f is a bijection, there exists some r₀ ∈ R such that f(r₀) = D.
However, by the definition of D, we have:
- If r₀ ∈ D, then r₀ ∉ f(r₀) = D, which is a contradiction.
- If r₀ ∉ D, then r₀ ∈ f(r₀) = D, which is also a contradiction.
Therefore, our assumption that there exists a bijective function f : R -> S
must be false.
Therefore, |S| > |R|.
- LegionMammal978 1y agoIf it's the idea of completed infinity that's the objection, then it's the first step, constructing the powerset, that would be problematic. Various forms of finitism would not accept that one can 'take all the subsets, finite or infinite, and quantify over them' and obtain a meaningful result past the formal level.
- drdeca 1y agoThey should still then accept that there is no surjection from a set to a set that is a set of all subsets of that set (because of there being no such set).
- LegionMammal978 1y agoThere would be no such surjection, but there would also be no such |S| to talk about in the conclusion.