# Question on singular cardinals

This is a question concerning the lemma discussed in Question about a proof about singular cardinals. In this question, it is proved that if an infinite cardinal $\kappa$ is singular, then it can be written as the sum of strictly smaller cardinals over a strictly smaller index set. What about the converse? If $\kappa=\sum_{\alpha<\lambda}\kappa_{\alpha}$, with $\lambda<\kappa$ and $\kappa_{\alpha}<\kappa$, then it follows easily that $\kappa=sup_{\alpha<\lambda}\kappa_{\alpha}$. But how do we get a strictly increasing sequence from this? Jech says by transfinite recursion, which I don't really know, so maybe there is a more elementary way of seeing it?

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Let $\mu_\gamma=\sum_{\alpha<\gamma}\kappa_\alpha$, clearly $\mu_\alpha$ is a non-decreasing sequence, let $\{\mu_\alpha\mid\alpha<\lambda\}$ be the set of those cardinals. It has a strictly increasing natural order as a set of ordinals, which we can use to generate a strictly increasing sequence.

It remains to point out that this set has size $\lambda$, but it cannot have more elements, and not less because $\lambda$ is the cofinality of $\kappa$.

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Given any cofinal function $f\colon \lambda\to\kappa$ it is a simple matter to derive a strictly increasing cofinal function $g\colon \lambda\to\kappa$ from it. This is defined as follows: $$g(\alpha)=\max\{f(\alpha),(\sup_{\beta<\alpha}g(\beta))+1\}$$ Note that this is really a definition by transfinite recursion, since we use the previous values of $g$ to define its value at $\alpha$. The $f(\alpha)$ term of the definition ensures that $g$ is cofinal and the $\sup_{\beta<\alpha} g(\beta)$ term ensures that it is strictly increasing.

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This doesn’t ensure that $g$ is strictly increasing. – Brian M. Scott Jan 17 '13 at 17:40
@BrianM.Scott You're right. I have a (bad) habit of ignoring finite quantities. What's +1 if there might be a measurable cardinal running around. – Miha Habič Jan 17 '13 at 17:49
Immeasurably tiny! – Brian M. Scott Jan 17 '13 at 17:59