[proofplan] We must show $f(\sup_i \alpha_i) = \sup_i f(\alpha_i)$. The argument splits on whether the collection $\{\alpha_i\}_{i \in I}$ has a maximum element. If it does, both sides reduce to $f$ applied at that maximum and equality is immediate from monotonicity. If it does not, the supremum $\alpha := \sup_i \alpha_i$ is a limit ordinal, and we exploit the continuity of $f$ at $\alpha$: the left-hand side unfolds as $\sup\{f(\gamma) : \gamma < \alpha\}$. Each $f(\alpha_i)$ sits in this set, giving one inequality; each $\gamma < \alpha$ is dominated by some $\alpha_i > \gamma$ (since no $\alpha_i$ is maximal), giving the reverse. Throughout we use that $f$ is strictly increasing, a consequence of normality already established in [Normal Functions are Strictly Increasing and Dominate the Identity](/theorems/1474). [/proofplan] [step:Dispose of the case where $\{\alpha_i\}$ has a maximum element] Suppose there exists $j \in I$ with $\alpha_j \geq \alpha_i$ for all $i \in I$. Then by definition of supremum, \begin{align*} \sup_{i \in I} \alpha_i = \alpha_j. \end{align*} By [strict monotonicity of $f$](/theorems/1474), $\alpha_i \leq \alpha_j$ implies $f(\alpha_i) \leq f(\alpha_j)$ for every $i \in I$. Hence $f(\alpha_j)$ is an upper bound for $\{f(\alpha_i)\}_{i \in I}$, and it belongs to this set (take $i = j$), so \begin{align*} \sup_{i \in I} f(\alpha_i) = f(\alpha_j) = f\!\left(\sup_{i \in I} \alpha_i\right), \end{align*} which is the desired equality. We henceforth assume $\{\alpha_i\}_{i \in I}$ has no maximum element. [/step] [step:Identify $\alpha := \sup_i \alpha_i$ as a limit ordinal] Set $\alpha := \sup_{i \in I} \alpha_i$. We claim $\alpha$ is a limit ordinal. The collection $\{\alpha_i\}_{i \in I}$ is non-empty by hypothesis, so $\alpha$ is a well-defined ordinal. Suppose, for contradiction, that $\alpha$ is a successor, $\alpha = \delta^+$. Then each $\alpha_i \leq \delta^+ = \alpha$, and since there is no maximum, strict inequality $\alpha_i < \alpha = \delta^+$ must hold for every $i \in I$. But $\alpha_i < \delta^+$ is equivalent to $\alpha_i \leq \delta$ on the ordinals, so $\delta$ is an upper bound for $\{\alpha_i\}$. This forces $\alpha = \sup_i \alpha_i \leq \delta < \delta^+ = \alpha$, a contradiction. If $\alpha = 0$, then $\alpha_i \leq 0$ and $\alpha_i = 0$ for all $i$, so $\{\alpha_i\}$ has maximum element $0$, contradicting our standing assumption. Hence $\alpha$ is neither zero nor a successor, so $\alpha$ is a limit ordinal. [guided] We are in the case where no $\alpha_j$ is maximal among the $\alpha_i$. We want to apply continuity of $f$ at $\alpha := \sup_i \alpha_i$, which requires knowing that $\alpha$ is a limit ordinal. Why is it? An ordinal is either zero, a successor, or a limit, so we rule out the first two. **Could $\alpha = 0$?** If so, then every $\alpha_i \leq 0$, meaning every $\alpha_i = 0$. But then $\alpha_i = 0$ is a maximum, contradicting the no-maximum hypothesis. **Could $\alpha = \delta^+$ be a successor?** The key ordinal fact is that $\mu < \delta^+ \iff \mu \leq \delta$. Since no $\alpha_i$ achieves the supremum $\alpha = \delta^+$ (otherwise $\alpha_i = \alpha$ would be a maximum), each $\alpha_i < \delta^+$, so each $\alpha_i \leq \delta$. This makes $\delta$ a smaller upper bound for $\{\alpha_i\}$, contradicting $\alpha = \sup_i \alpha_i = \delta^+ > \delta$. By elimination, $\alpha$ is a limit ordinal. This is the property we need to unlock the continuity definition of "normal" at $\alpha$. [/guided] [/step] [step:Expand $f(\alpha)$ using continuity and show $\sup_i f(\alpha_i) \leq f(\alpha)$] Since $\alpha$ is a limit ordinal, the continuity property of the normal function $f$ gives \begin{align*} f(\alpha) = \sup\{f(\gamma) : \gamma < \alpha\}. \end{align*} For each $i \in I$, we have $\alpha_i \leq \alpha$, and by [strict monotonicity of $f$](/theorems/1474), $f(\alpha_i) \leq f(\alpha)$. Taking the supremum over $i$: \begin{align*} \sup_{i \in I} f(\alpha_i) \leq f(\alpha). \end{align*} [/step] [step:Prove the reverse inequality $f(\alpha) \leq \sup_i f(\alpha_i)$ by dominating each $\gamma < \alpha$] We show that every term $f(\gamma)$ appearing in the supremum $f(\alpha) = \sup\{f(\gamma) : \gamma < \alpha\}$ is bounded above by some $f(\alpha_i)$. Let $\gamma < \alpha = \sup_i \alpha_i$. Since $\alpha$ is the *least* upper bound of $\{\alpha_i\}$, $\gamma$ cannot be an upper bound: otherwise $\alpha \leq \gamma < \alpha$, a contradiction. Hence there exists $i \in I$ with $\alpha_i > \gamma$. By [strict monotonicity of $f$](/theorems/1474), \begin{align*} f(\gamma) < f(\alpha_i) \leq \sup_{j \in I} f(\alpha_j). \end{align*} Thus every $f(\gamma)$ with $\gamma < \alpha$ satisfies $f(\gamma) \leq \sup_j f(\alpha_j)$, so $\sup_j f(\alpha_j)$ is an upper bound for $\{f(\gamma) : \gamma < \alpha\}$. Taking the supremum: \begin{align*} f(\alpha) = \sup\{f(\gamma) : \gamma < \alpha\} \leq \sup_{j \in I} f(\alpha_j). \end{align*} [guided] Continuity unfolds $f(\alpha)$ as $\sup\{f(\gamma) : \gamma < \alpha\}$ — a supremum over **all** ordinals below $\alpha$, not just the indexed family $\{\alpha_i\}$. The indexed family is generally much smaller than this set. Why is the supremum over the bigger set nevertheless bounded by the supremum over the smaller indexed set? The bridge is the definition of $\alpha$ as the *least* upper bound of $\{\alpha_i\}$. Concretely: for any $\gamma < \alpha$, $\gamma$ is strictly less than the least upper bound of $\{\alpha_i\}$, so $\gamma$ is not itself an upper bound. Hence there must be some $\alpha_i > \gamma$ — otherwise $\gamma$ would be an upper bound smaller than $\alpha$, contradicting minimality of $\alpha$ as an upper bound. Once we have $\alpha_i > \gamma$, strict monotonicity of $f$ gives $f(\gamma) < f(\alpha_i) \leq \sup_j f(\alpha_j)$. This argument works because we are evaluating $f$ at every $\gamma < \alpha$ in a set that is *cofinal* in the $\alpha_i$: the indexed family is "close enough" to $\alpha$ to dominate every sub-$\alpha$ witness. Taking the supremum over $\gamma < \alpha$: \begin{align*} f(\alpha) = \sup\{f(\gamma) : \gamma < \alpha\} \leq \sup_{j \in I} f(\alpha_j). \end{align*} [/guided] [/step] [step:Combine the two inequalities to conclude equality] From the previous two steps, \begin{align*} \sup_{i \in I} f(\alpha_i) \leq f(\alpha) \leq \sup_{j \in I} f(\alpha_j). \end{align*} Both outer terms are equal, so all three are equal. Substituting $\alpha = \sup_i \alpha_i$: \begin{align*} f\!\left(\sup_{i \in I} \alpha_i\right) = \sup_{i \in I} f(\alpha_i). \end{align*} This completes the proof. [/step]