Theorem:A set S is infinite if and only if for all n∈ℕ, there exists a subset of S whose cardinality is n.
I think their necessary condition $\rightarrow$ proof is correct. I won't add their forwards proof because it seems tedtious.
Proof I found this from proof wiki
and it's wrong I think. I'm looking to confirm that their proof is wrong. I wrote my own and wanted proof verification.
$\leftarrow$
Sufficient Condition Suppose that for all n∈ℕ, there exists a subset of S whose cardinality is n.
Assume that S is finite.
Let N=|S|.
As N∈ℕ it follows that N+1∈ℕ.
By hypothesis, there exists a subset T⊆S whose cardinality is N+1.
From Cardinality of Subset of Finite Set, |S|≥|T|.
But then |S|=N≥N+1=|T|, which contradicts the fact that N
From this contradiction it follows that S can not be finite.
I disagree that T, a proper subset of S, has a cardinality greater than S. By definition, a proper subset cannot have a cardinality greater or equal to that set that it is a proper subset of, that is Card T
I disagree that by hypothesis, $\exists$ a subset T $\subseteq$ S whose cardinality is N+1. Card T should be less than S. since T is subset of S.
I think their proof by contradiction is a mistake and direct proof is better.
I think a better way is to
Pf. Assume for all n∈ℕ, there exists a proper subset of S whose cardinality is n. Try to show that S is infinite by induction.
Base case is the case when the proper subset of S called T is of cardinality 1. If you think the natural numbers start at 0, then consider the empty set which is a subset of every set and Card 0 + Card $S\setminus \emptyset$ = Card S.
If you think natural numbers start at 1 then,
Card $S\setminus T$ + Card T = Card S.
Inductive step: Assume for all sets with cardinality denoted by the number 1<=j<=k where j is the cardinality of T the equation Card $S\setminus T$ + Card T = Card S holds. Show this equation works for k+1 being the cardinality of set T.
When we add an element to set T, which is a proper subset of S Card T is increased by one. the cardinality of $S\setminus T$ decreases by one, so we end up adding and subtracting one making no net difference. the equation holds for when k+1 is the cardinality of T.
For inductive step:
(Card $S\setminus T$) $-1$ + Card T +1 = Card S
$\Longleftrightarrow$
$x$ $\in$ S.
Card $((S\setminus T) \setminus {{x}})$ + Card (T $\cup$ {x}) = Card S
Therefore S is infinite.
Right, that's why it's a proof by contradiction. The original assumption lead to a conclusion that is clearly false, hence the assumption is false. It simply is the case that, by hypothesis, $S$ has a subset of cardinality $N+1$. For it is a proeprty of $S$ that it has a subset of cardinality $n$ for any $n\in\mathbb N$. Since $N+1\in\mathbb N$, $S$ has a subset of cardinality $N+1$. There's no flaw in this reasoning. It's also true that we need $\vert T \vert \leq \vert S \vert$, and so we reach a contradiction, meaning that $S$ must be infinite.
Further, proving the result by induction doesn't really work, since your goal isn't to prove that a certain statement holds for all natural numbers.