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Can someone give me intuitive explanation of cofinality?

I pretty much understand what does it mean for a subset $x$ in $y$ to be cofinal (something like being dense when going from the left to the right), but when defining cofinality of $y$, one first says that

$$\text{cof}(y) = \text{min}\left\{\text{otp}(x) | x \text{ cofinal in } y\right\}$$

On the other hand, one states that this is the same as if we take $\text{card}(x)$ instead of $\text{otp}(x)$ in the above definition, but I wonder why..

Asaf Karagila
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edward_scissorhands
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  • It's always good to give a context at the start of your question. "Cofinality" is a term used in a several different contexts. – Thomas Andrews Jan 11 '17 at 19:52
  • In may case it concerns set theoretical approach. Namely, I've read the next: To get some more information on cardinal exponentiation, we need to measure how “fast” a cardinal can be approximated using smaller cardinals. Definitions: a) A set x ⊆ λ is cofinal in the limit ordinal λ if ∀α < λ ∃ξ ∈ x α < ξ . b) The cofinality of a limit ordinal λ is cof(λ) = min {otp(x)|x ⊆ λ is cofinal in λ}. c) A limit ordinal λ is regular if cof(λ) = λ ; otherwise λ is singular. But still I have problem with intuitive understanding and especially with the above equation that I've mentioned. – edward_scissorhands Jan 11 '17 at 20:00

1 Answers1

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The point is that a cofinality is always a cardinal. In other words, if $\alpha$ is an ordinal which is not a cardinal, then $\operatorname{cf}(\alpha)<\alpha$.

To see why, enumerate $\alpha$ by its cardinality, and start constructing a strictly increasing sequence which agrees on both the enumeration and the natural ordering of $\alpha$. You either get stuck at some ordinal below $|\alpha|$, or that you get that there is a cofinal sequence of length and order type $|\alpha|$.

If the cofinal sequence you ended up with was not of a cardinal order type, repeat the process. This is a decreasing sequence of ordinals, so it has to stop: exactly when the cofinal sequence is a cardinal, and it is not hard to check that this sequence is indeed cofinal in $\alpha$.

Note that it is often the case that "the smallest ..." ends up with a cardinal. It's not always, sure, but it isn't peculiar for cofinality. Other definitions include the Hartogs and Lindenbaum numbers.

Asaf Karagila
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  • http://math.stackexchange.com/questions/24982/cofinality-and-its-consequences/ and http://math.stackexchange.com/questions/389420/why-does-every-countable-limit-ordinal-have-cofinality-omega/ might be relevant. – Asaf Karagila Jan 11 '17 at 20:23
  • Out of curiosity: What are Lindenbaum numbers? Google doesn't tell me :) – abc Jan 12 '17 at 19:10
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    @Achilles: If $x$ is a set, the Lindenbaum number of $x$, $\aleph^*(x)$, is the least ordinal such that there is no surjection from $x$ onto it. – Asaf Karagila Jan 12 '17 at 21:46