A Successor Cardinal is Regular

Question

Trying to show that every cardinal $k$ , $k^+$ , its successor, is regular. This is what I've come up with. Thoughts?

If this does not hold, then a cofinal map $f: \lambda\rightarrow k$ where $\lambda\leq k$ would exist. This would mean $k^+$ would be a $k$ union of sets each of size $\leq k$ . This would contradict the following:

Let $k \in CARD$ and let $X=\cup_{\alpha<k}X_\alpha$ where $|X_\alpha|\leq k$ for all $\alpha < k$ . Then $|X|\leq k$ [proof: Using the Axiom of Choice, let $f$ be a function with domain$k$ such that for all $\alpha<k$ ,$f(\alpha):k\rightarrow X_\alpha$ is a bijection. Define $F: k \times k\rightarrow X$ $F(\alpha,\beta)=f(\alpha)(\beta)$ . Clearly $F$ is onto and the result follows by the Fundamental theorem of cardinal arithmetic]

Answer 1

The proof is okay, but you can make it ever so simpler by noting that:

1. If $f\colon\kappa\to\kappa^+$, then for every $\alpha$, $f(\alpha)$ has cardinality at most $\kappa$,
2. The supremum of a set of ordinal is its union.
3. So $\sup\operatorname{rng}f$ is the union of at most $\kappa$ sets of size at most $\kappa$,and now appeal to choice and cardinal arithmetic.

I should also mention that it is consistent that $\omega_1$ is singular, so the use of choice is indispensable.