[proofplan] We prove each assertion directly from the definition of cofinality. For a successor ordinal, the predecessor alone is cofinal. For a nonzero limit ordinal, every finite subset is bounded below $\alpha$, so no finite cofinal subset exists. Finally, a cofinal increasing map $f: \delta \to \alpha$ supplies a cofinal subset of $\alpha$ of cardinality at most $|\delta|$. [/proofplan] [step:Use the predecessor as a cofinal subset of a successor ordinal] Assume that $\alpha = \beta + 1$ for an ordinal $\beta$. Define the subset $C \subset \alpha$ by $C := \{\beta\}$. Since $\beta < \beta + 1 = \alpha$, this is indeed a subset of $\alpha$. To check that $C$ is cofinal in $\alpha$, let $\gamma < \alpha$. Since $\alpha = \beta + 1$, every ordinal below $\alpha$ is at most $\beta$, so $\gamma \leq \beta$. Hence $C$ is cofinal in $\alpha$. Therefore \begin{align*} \operatorname{cf}(\alpha) \leq |C| = 1. \end{align*} Since $\alpha$ is nonzero, no empty subset of $\alpha$ is cofinal in $\alpha$. Thus $\operatorname{cf}(\alpha) \neq 0$, and consequently $\operatorname{cf}(\alpha) = 1$. [/step] [step:Show that no finite subset is cofinal in a nonzero limit ordinal] Assume that $\alpha$ is a nonzero limit ordinal. By definition, $\operatorname{cf}(\alpha)$ is a cardinal, so it remains to prove that it is not finite. Let $F \subset \alpha$ be a finite subset. If $F = \varnothing$, then $F$ is not cofinal in $\alpha$ because $\alpha \neq 0$, so there exists an ordinal $\gamma < \alpha$, for instance $\gamma = 0$, with no element of $F$ above it. Now suppose $F \neq \varnothing$. Since ordinals are linearly ordered by membership and $F$ is finite, $F$ has a maximum element; denote it by $\mu \in F$. Since $F \subset \alpha$, we have $\mu < \alpha$. Because $\alpha$ is a limit ordinal, it is not equal to $\mu + 1$, and therefore \begin{align*} \mu + 1 < \alpha. \end{align*} Every element of $F$ is at most $\mu$, so no element of $F$ is greater than or equal to $\mu + 1$. Hence $F$ is not cofinal in $\alpha$. Thus no finite subset of $\alpha$ is cofinal in $\alpha$. Since $\alpha$ itself is a cofinal subset of $\alpha$, cofinal subsets exist, and the least possible cardinality of such a subset is an infinite cardinal. Therefore $\operatorname{cf}(\alpha)$ is an infinite cardinal. [guided] Assume that $\alpha$ is a nonzero limit ordinal. The cofinality $\operatorname{cf}(\alpha)$ is defined as the least cardinality of a cofinal subset of $\alpha$, so it is automatically a cardinal once we know that cofinal subsets exist. They do exist: the whole set $\alpha$ is cofinal in itself, because every $\gamma < \alpha$ is bounded above by itself, an element of $\alpha$. It remains to rule out finite cofinal subsets. Let $F \subset \alpha$ be finite. If $F = \varnothing$, then $F$ cannot be cofinal in $\alpha$: since $\alpha \neq 0$, the ordinal $0$ satisfies $0 < \alpha$, and there is no element of the empty set above $0$. Now suppose $F \neq \varnothing$. A finite nonempty set of ordinals has a largest element because ordinals are linearly ordered. Let $\mu \in F$ denote this maximum. Since $F \subset \alpha$, we have $\mu < \alpha$. The limit ordinal hypothesis is used exactly here: because $\alpha$ is a limit ordinal, no successor ordinal reaches $\alpha$ from below. Hence \begin{align*} \mu + 1 < \alpha. \end{align*} But every element of $F$ is at most $\mu$, so no element of $F$ is at least $\mu + 1$. Therefore $F$ fails the defining condition for cofinality. We have shown that every finite subset of $\alpha$ is not cofinal. Since cofinal subsets exist and the cofinality is the least cardinality of such a subset, $\operatorname{cf}(\alpha)$ cannot be finite. Therefore $\operatorname{cf}(\alpha)$ is an infinite cardinal. [/guided] [/step] [step:Bound cofinality by the size of any cofinal indexing ordinal] Let $\delta$ be an ordinal, and let $f: \delta \to \alpha$ be an increasing map whose range is cofinal in $\alpha$. Define the range subset $R \subset \alpha$ by \begin{align*} R := f[\delta] = \{f(\xi) : \xi < \delta\}. \end{align*} By hypothesis, $R$ is cofinal in $\alpha$. Therefore, by the definition of cofinality, \begin{align*} \operatorname{cf}(\alpha) \leq |R|. \end{align*} The map \begin{align*} f: \delta &\to R \\ \xi &\mapsto f(\xi) \end{align*} is surjective by the definition of $R$, so $|R| \leq |\delta|$. Combining the two inequalities gives \begin{align*} \operatorname{cf}(\alpha) \leq |R| \leq |\delta|. \end{align*} Thus $\operatorname{cf}(\alpha) \leq |\delta|$. [/step]