Solution verification: proof that the supremum of a set of cardinals is a cardinal

Question

I'm trying to show that if $X$ is a set of cardinals, then $\sup X$ is a cardinal.

This is part (ii) of Lemma 3.4 in the third edition of Jech on page 29.

The proof in Jech is pretty terse and I don't really understand it.

Here's the proof in Jech.

Let $\alpha = \sup X$. If $f$ is a one-to-one mapping of $\alpha$ onto some $\beta < \alpha$, let $\kappa \in X$ be such that $\beta < \kappa \le \alpha$. Then $\left|\kappa\right| = \left|\left\{ f(\xi) : \xi < \kappa \right\}\right| \le \beta$, a contradiction. Thus $\alpha$ is a cardinal.

I have read this question and this question, but I still don't really understand the standard proof. (I'm writing this with the assumption that Jech and those two questions are all referencing a proof that's essentially the same)

I attempted to prove this fact myself contrapositively below and I have two questions about it:

1. Is it valid?

2. Does it differ substantially from the standard proof?

---

Construe $X$ as a set of ordinals.

By way of contrapositive, we want to show that if $\sup X$ is a non-cardinal ordinal (NCO), then X contains at least one NCO.

Let $\kappa$ be $\left|\sup X\right|$.

Observe that $X$ does not contain any $\alpha$ where $\left|\alpha\right| > \kappa$, if it did $\sup X$ would have a cardinality that's too great.

Let $X'$ be $\{ \alpha \in X : \left|\alpha\right| = \kappa \}$.

If $X'$ is empty, then $\sup X \le \kappa$, which is a contradiction.

Suppose $X'$ is a singleton. If $\cup X'$ is a cardinal, then $\sup X$ is a cardinal, which is a contradiction. If $\cup X'$ is a NCO, then $X$ contains a NCO as desired.

Suppose $X'$ contains at least two elements. Then at least one element is a NCO, so $X$ contains at least one NCO.

Answer 1

I'll break down Jech's proof, as once you understand his proof I believe your question will be answered.

Let Card be the Proper class of all cardinals, and X be a set.

Theorem:

If X $\subset$ Card, then sup(X)$\in$Card

Proof:

Assume

X $\subset$ Card

Remark: by a similar theorem ( which you should prove)

If X $\subset$ Ord then sup(X)$\in$Ord

Thus, sup(X)=α$\in$Ord

We will show that α is also a Cardinal.

Assume (RAA) that α is not a cardinal,

So $\exists$β$\in$α(β ≈ α)

i.e. there exists a bijection from α to β

Let f:α $\to$ β, be such a bijection

As α is the least upper bound of X and β < α

β is not an upper bound of X

Thus we can find κ$\in$X s.t.

β < κ ≤ α

As κ is a Cardinal |κ| > |β|

But

rang(f) = β

and

rang($f\restriction_κ$) = { f(δ) | δ$\in$κ} $\subset$ rang(f)

Notice that |κ| = |rang($f\restriction_κ$)| ≤ |rang(f)|

And so κ ≤ |β| < κ

κ < κ

A contradiction

Answer 2

Suppose $\alpha$ is not a cardinal. Then $\alpha$ is equinumerous to some $\beta<\alpha$. Let $\gamma \in X$ be such that $\beta<\gamma$. Then

$$|\alpha|=|\beta|\leq \beta <\gamma =|\gamma|\leq |\alpha|$$

a contradiction.