Is every function from $\aleph_0 \to \aleph_2$ bounded?

Question

If $f$ is a function $f:\aleph_0 \to \aleph_2$, does it mean that the range of f is bounded in $\aleph_2$? Does this hold for all regular cardinals?

Well, it doesn't quite hold for all regular cardinals in place of $\aleph_2$, just uncountable ones :-) — Trevor Wilson, Nov 18 '13 at 18:46

score 5 · Answer 1 · edited Apr 13 '17 at 12:58

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Yes. Recall that given a cardinal $\kappa$ the cofinality of $\kappa$, $\mathrm{cf} ( \kappa )$, is the least cardinal $\mu$ for which there is an unbounded (cofinal) function $\mu \to \kappa$. Regularity means that $\mathrm{cf} ( \kappa ) = \kappa$, and all successor cardinals are regular.

Added to indoctrinate others into Asaf's anti-Choice programme

Note, also, that there is a fair bit of Choice involved in the above. Without the Axiom of Choice it is possible that that $\aleph_2$ is singular; in particular it could have cofinality $\omega$, meaning that there is an unbounded function $\aleph_0 \to \aleph_2$. The answers to this old math.SE question and well as this slightly more recent MO question contain a wealth of information on the broader topic.

edited Apr 13 '17 at 12:58

Community

1

answered Nov 18 '13 at 17:29

user642796

52,188

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Seeing there is an abundance of answers, I feel there's no need for another answer. I expect you to be the one to add a line about the axiom of choice. – Asaf Karagila Nov 18 '13 at 17:45
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I have a programme now? Cool! – Asaf Karagila Nov 18 '13 at 18:35
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(Let me add that, even without choice, if $f:\omega\to\omega_2$ is cofinal, then for some $n$, the gap between $f(n)$ and $f(n+1)$ must be uncountable.) – Andrés E. Caicedo Nov 18 '13 at 23:49

score 3 · Answer 2 · answered Nov 18 '13 at 17:29

If $\kappa$ is a regular cardinal, and $\alpha$ is any ordinal less than $\kappa$, then every function $f:\alpha\to\kappa$ is bounded. In particular, any function $f:\omega\to\omega_2$ is bounded.

More generally, all functions $f:\alpha\to\kappa$ will be bounded if $\alpha<\operatorname{cf}\kappa$, the cofinality of $\kappa$.

score 3 · Answer 3 · answered Nov 18 '13 at 17:30

One of the definitions of a cardinal $\kappa$ being regular is that, whenever $\alpha < \kappa$, every function $f : \alpha \to \kappa$ is bounded.

In any case, you can prove this directly, using the fact that a countable union of sets of cardinality $\aleph_1$ has cardinality $\aleph_1$: consider $$\bigcup_{n < \omega} \{ \alpha < \omega_2 : \alpha \le f(n) \}$$

Is every function from $\aleph_0 \to \aleph_2$ bounded?

3 Answers3