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.

Now I wonder why does $cof(κ) > λ$ imply that every function from $λ$ into $κ$ is bounded by some ordinal $ν < κ$, where $λ ∈ Card$ and $κ > λ$

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    $\begingroup$ What can be the order type of the range of such function? $\endgroup$ – nombre Jan 12 '17 at 12:34
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    $\begingroup$ Oh actually you probably need an intermediate result to prove this, namely, that every regular ordinal is a cardinal. (i.e. is initial) $\endgroup$ – nombre Jan 12 '17 at 12:43
  • $\begingroup$ @user254665: If it is any function, it sure can. For instance there are surjective maps $\omega_0 \rightarrow \omega_0+1$. $\endgroup$ – nombre Jan 12 '17 at 12:49
  • $\begingroup$ @nombre. I meant to say something else . Sorry . $\endgroup$ – DanielWainfleet Jan 12 '17 at 12:53
  • $\begingroup$ It is any function, but still I don't get why $\endgroup$ – edward_scissorhands Jan 12 '17 at 13:18

For any ordinal $\lambda$, $cof(\lambda)$ is the least cardinal $Cf(\lambda)$ of a cofinal subset of $\lambda$.

Indeed, by definition, since for any ordinal $\alpha$, $|\alpha| \leq \alpha$, $cof(\lambda) \leq Cf(\lambda)$.

If $\lambda$ has a greatest element, both of them are $1$, and if $\lambda$ is empty, both of them are $0$, so in the following I assume $\lambda$ is limit non zero.

Now let $C$ be a cofinal subset of $\lambda$ of cardinal $Cf(\lambda)$. Fix a surjection $c: Cf(\lambda) \rightarrow C$. We want to define a subset $D:=\{d_{\alpha} \ | \ \alpha < Cf(\lambda)\}$ of $\lambda$ order type $Cf(\lambda)$ which is cofinal in $\lambda$. This can be done by induction: if the $d_{\beta}$ are defined below a certain ordinal $\alpha < Cf(\lambda)$, then by minimality of $Cf(\lambda)$, $\{d_{\beta} \ |\ \beta < \alpha\}$ has an upper bound in $\lambda$, and since $\lambda$ doesn't have a greatest element, it has a strict upper bound $\delta$, we define $d_{\alpha} := \max(\delta, c_{\alpha})$. By construction, $\alpha \mapsto c_{\alpha}$ is strictly increasing (because $d_{\alpha} \geq \delta$) and cofinal (because $d_{\alpha} \geq c_{\alpha}$ and $C$ is cofinal). So $Cf(\lambda) \leq cof(\lambda)$.

This anwsers your question: for $f: \lambda \rightarrow \kappa$, $|f(\lambda)| \leq \lambda < cof(\kappa) = Cf(\kappa)$ so $f(\lambda)$ is not cofinal in $\kappa$.

There are several important and basic properties of the cofinality / of regular ordinals which I suggest you look into:

-every regular ordinal is a cardinal (this follows easily from the result above)

-every successor infinite cardinal is regular

-the cofinality of an ordinal is regular

-the cofinality of an ordinal is the only regular ordinal among the order types of cofinal subsets of that ordinal


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