[proofplan] We reduce the problem to the well-ordering of a single initial segment. Pick any $\alpha \in S$. Either $\alpha$ is already the least element, or some element of $S$ lies strictly below $\alpha$, i.e. in the initial segment $I_\alpha = \{\beta : \beta < \alpha\}$. Because $I_\alpha$ is well-ordered by $<$ (every ordinal is a well-ordered set under its own order), the non-empty subset $S \cap I_\alpha$ contains a least element $\beta$, which we show is the least element of the whole of $S$. [/proofplan] [step:Fix an element $\alpha \in S$ and dispose of the trivial case] Since $S$ is non-empty, fix any $\alpha \in S$. Consider the set \begin{align*} T := S \cap I_\alpha = \{\xi \in S : \xi < \alpha\}. \end{align*} If $T = \varnothing$, then no element of $S$ is strictly less than $\alpha$, so for every $\xi \in S$ we have $\xi \geq \alpha$. Hence $\alpha$ is the least element of $S$ and we are done. Assume from now on that $T \neq \varnothing$. [guided] The idea is to eliminate the easy case before doing any real work. Given our choice $\alpha \in S$, there are two possibilities. *Case 1: Nothing in $S$ lies below $\alpha$.* Then $\alpha$ is automatically the minimum. *Case 2: Something in $S$ lies below $\alpha$.* Then we must find the least such element. Formally, the "things in $S$ below $\alpha$" form the set $T := S \cap I_\alpha$, where $I_\alpha := \{\beta : \beta < \alpha\}$ is the initial segment of ordinals strictly below $\alpha$. If $T = \varnothing$, then no $\xi \in S$ satisfies $\xi < \alpha$; by [trichotomy of ordinal comparison](/theorems/1467), this forces $\xi \geq \alpha$ for every $\xi \in S$, so $\alpha = \min S$. The interesting case, which we handle next, is $T \neq \varnothing$. [/guided] [/step] [step:Apply the well-ordering of the initial segment $I_\alpha$ to extract the least element of $T$] The initial segment $I_\alpha = \{\beta : \beta < \alpha\}$ is itself an ordinal (namely, $\alpha$ itself under the identification of an ordinal with its set of predecessors), hence well-ordered by $<$. Since $T \subseteq I_\alpha$ is non-empty, the [well-ordering property](/theorems/1470) of $I_\alpha$ produces a least element \begin{align*} \beta := \min T \in T. \end{align*} By construction $\beta \in S$ and $\beta < \alpha$. [guided] Why do we get a least element of $T$ for free here? The answer is the **definition of an ordinal**: an ordinal $\alpha$ is a transitive set well-ordered by $\in$ (equivalently, by $<$). "Well-ordered" means: every non-empty subset of $\alpha$ has a least element. The initial segment $I_\alpha$ **is** $\alpha$ (viewed as its own set of predecessors), so any non-empty subset of $I_\alpha$ has a least element. We verify the hypothesis of the well-ordering property. We need a non-empty subset of $I_\alpha$. Our subset is $T = S \cap I_\alpha$. It is a subset of $I_\alpha$ by construction, and it is non-empty by the assumption we made at the end of the previous step. Hence $\beta := \min T$ exists, and this $\beta$ satisfies: \begin{align*} \beta \in T \subseteq S, \qquad \beta < \alpha, \qquad \forall\, \xi \in T: \beta \leq \xi. \end{align*} The first two facts say that $\beta$ is an element of $S$ that lies below $\alpha$. The third says $\beta$ is the least such. It remains to show $\beta$ is the least element not just of $T$, but of all of $S$. [/guided] [/step] [step:Show $\beta$ is a lower bound for all of $S$] We claim $\beta \leq \xi$ for every $\xi \in S$. Fix $\xi \in S$ and consider the two cases given by [trichotomy for ordinals](/theorems/1467): *Case 1: $\xi < \alpha$.* Then $\xi \in S \cap I_\alpha = T$, so $\beta \leq \xi$ by minimality of $\beta$ in $T$. *Case 2: $\xi \geq \alpha$.* Since $\beta < \alpha$ from the previous step, transitivity of $<$ on ordinals gives $\beta < \alpha \leq \xi$, hence $\beta \leq \xi$. Both cases yield $\beta \leq \xi$. Combined with $\beta \in S$ from the previous step, this shows $\beta = \min S$, completing the proof. [guided] We established that $\beta$ is the least element of $T = S \cap I_\alpha$. But $T$ only contains the elements of $S$ below $\alpha$ — what about the elements of $S$ at or above $\alpha$? We need to show $\beta$ is still below those too. Pick an arbitrary $\xi \in S$. By trichotomy for ordinals, exactly one of $\xi < \alpha$, $\xi = \alpha$, $\xi > \alpha$ holds. Combining the last two as $\xi \geq \alpha$, we split: *Case 1: $\xi < \alpha$.* Then $\xi \in I_\alpha$, and since $\xi \in S$ as well, $\xi \in S \cap I_\alpha = T$. By the defining property of $\beta = \min T$, we get $\beta \leq \xi$. *Case 2: $\xi \geq \alpha$.* Here $\xi$ lives outside $T$. But recall $\beta < \alpha$. Chaining, \begin{align*} \beta < \alpha \leq \xi, \end{align*} which gives $\beta < \xi$, so certainly $\beta \leq \xi$. In both cases $\beta \leq \xi$. Since this holds for all $\xi \in S$ and $\beta \in S$, we conclude $\beta = \min S$. The set $S$ has a least element, as claimed. [/guided] [/step]