# Cofinality and partitions of a cardinal

Given a (limit) ordinal $\alpha$ we could define its cofinality $\text{cf}(\alpha)$ as the least ordinal $\beta$ such that there's an unbounded function from $\beta$ to $\alpha$. It's not difficult to prove that $\text{cf}(\alpha)$ is a cardinal. However, for $\alpha$ cardinal we could define its cofinality to be the minimum cardinal $\beta$ such that there's a partition of $\alpha$ in $\beta$ pieces of strictly less cardinality than $\alpha$.

Why these definitions coincide? How do you go from one to the other when $\alpha$ is a cardinal?

## 2 Answers

Suppose $\alpha$ is a cardinal.

Suppose $\beta\geq cf(\alpha)$, and let $f: \beta \to \alpha$ be unbounded. Then $\alpha = \displaystyle\bigcup_{\delta\in \beta}f(\delta) = |\displaystyle\bigcup_{\delta\in \beta} f(\delta) | \leq \displaystyle\sum_{\delta \in \beta} |f(\delta)|$.

For $\delta \in \beta$, $f(\delta) <\alpha$ and since $\alpha$ is a cardinal each of the $f(\delta)$ is of cardinality strictly less than $\alpha$ so we have found a partition as desired.

Converseley, suppose $\alpha = \displaystyle\sum_{\delta \in \beta}\kappa_\delta$ is a partition and $\kappa_\delta$ is a cardinal strictly less than $\alpha$ and $\beta<\alpha$ ( we may assume $\beta <\alpha$ otherwise $\beta \geq \alpha$ and $\beta \geq cf(\alpha)$ follows immediately. We may also assume $\beta$ is a cardinal)

Consider then the family $\beta \to \alpha$, $\delta \mapsto \kappa_\delta$. If it were bounded below $\alpha$, say by $\mu$, then the cardinal of $\kappa_\delta$ would be $\leq |\mu|$, and the cardinal of the sum would be bounded by $\beta|\mu| = \max \{\beta, |\mu|\} <\alpha$, a contradiction. So this family is unbounded, and so $\beta \geq cf(\alpha)$, as desired.

We proved $\beta \geq cf(\alpha)$ if and only if there exists a partition of $\alpha$ in $\beta$ parts of cardinality $<\alpha$, so $cf(\alpha)$ is the least cardinal with this property.

Since we are talking about partitions, the order type matters less, so might as well assume $\alpha$ is a cardinal, since it makes things simpler.

Now given a partition of $\alpha$, such that all the parts are small, this means that each part is a bounded set. So fixing a minimal order type of a well ordering of the partition (that's $\beta$) we redefine each part to be an interval on $\alpha$. This is done recursively, of course, and the recursion can continue as long as we haven't reached $\alpha$ with our intervals. If we have then either not all parts are small, or $\beta$ isn't the least size of a partition.

Now the end points of the intervals give us an unbounded function from $\beta$ to $\alpha$. The other direction is essentially the same, going backwards.