[proofplan] We first build, from the set $A$ itself, the set of all ordinal order types of well-ordered subsets of $A$. Its union is an ordinal $\theta$, and the successor ordinal $\theta + 1$ cannot inject into $A$; otherwise its image would be a well-ordered subset of $A$ with order type $\theta + 1$, contradicting the definition of $\theta$. Among all ordinals that do not inject into $A$, we then choose the least one, call it $\kappa$, and use its minimality to show that every smaller ordinal injects into $A$. This rules out any bijection between $\kappa$ and a smaller ordinal, so $\kappa$ is initial. [/proofplan] [step:Construct an ordinal that cannot inject into $A$] Let $\mathcal{W}$ denote the set of all pairs $(B,R)$ such that $B \subset A$ and $R \subset B \times B$ is a well-ordering relation on $B$. Formally, \begin{align*} \mathcal{W} := \{(B,R) \in \mathcal{P}(A) \times \mathcal{P}(A \times A) : R \text{ well-orders } B\}. \end{align*} For each $(B,R) \in \mathcal{W}$, let $\operatorname{ot}(B,R)$ denote the unique ordinal order-isomorphic to the well-ordered set $(B,R)$. By Replacement applied to the map $(B,R) \mapsto \operatorname{ot}(B,R)$, the collection \begin{align*} T := \{\operatorname{ot}(B,R) : (B,R) \in \mathcal{W}\} \end{align*} is a set of ordinals. Define \begin{align*} \theta := \bigcup T. \end{align*} Since $T$ is a set of ordinals, $\theta$ is an ordinal. Let $\alpha := \theta + 1$ be the successor ordinal of $\theta$. We claim that no injective map $\alpha \to A$ exists. Suppose, for contradiction, that \begin{align*} f: \alpha \to A \end{align*} is injective. Define $B_f := f[\alpha] \subset A$, and define a relation $R_f \subset B_f \times B_f$ by \begin{align*} y_0 \, R_f \, y_1 \quad \Longleftrightarrow \quad f^{-1}(y_0) \in f^{-1}(y_1) \end{align*} for $y_0,y_1 \in B_f$. Because $f$ is injective, each element of $B_f$ has a unique preimage under $f$, so $R_f$ is well-defined. The map \begin{align*} f: (\alpha,\in) \to (B_f,R_f) \end{align*} is an order isomorphism, so $(B_f,R_f) \in \mathcal{W}$ and $\operatorname{ot}(B_f,R_f)=\alpha$. Hence $\alpha \in T$, and therefore $\alpha \subset \theta = \bigcup T$. This gives $\theta + 1 = \alpha \leq \theta$, impossible for ordinals. Thus no injective map $\alpha \to A$ exists. [/step] [step:Choose the least ordinal that does not inject into $A$] Using the ordinal $\alpha$ constructed in the previous step, define \begin{align*} S := \{\gamma \in \alpha + 1 : \text{there is no injective map } \gamma \to A\}. \end{align*} The set $S$ is nonempty because $\alpha \in S$. Since $\alpha + 1$ is well-ordered by membership, the nonempty subset $S$ has a least element. Let $\kappa$ denote this least element. Then $\kappa$ is an ordinal, no injective map $\kappa \to A$ exists, and for every ordinal $\beta < \kappa$, there exists an injective map $\iota_\beta: \beta \to A$; otherwise $\beta$ would belong to $S$, contradicting the minimality of $\kappa$. [/step] [step:Show that no smaller ordinal is equipotent to $\kappa$] Two sets are equipotent when there exists a bijection between them. Let $\beta$ be an ordinal with $\beta < \kappa$. Suppose, for contradiction, that $\beta$ is equipotent to $\kappa$. Then there exists a bijective map \begin{align*} b: \kappa \to \beta. \end{align*} By the minimality of $\kappa$, there exists an injective map \begin{align*} \iota_\beta: \beta \to A. \end{align*} The composition \begin{align*} \iota_\beta \circ b: \kappa \to A \end{align*} is injective, because a composition of injective maps is injective. This contradicts the defining property of $\kappa$, namely that no injective map $\kappa \to A$ exists. Therefore no ordinal $\beta < \kappa$ is equipotent to $\kappa$. [/step] [step:Conclude that $\kappa$ is initial and bounds $A$] An ordinal is initial precisely when it is not equipotent to any smaller ordinal. The previous step proves that no ordinal $\beta < \kappa$ is equipotent to $\kappa$, so $\kappa$ is an initial ordinal. By construction, no injective map $f: \kappa \to A$ exists. Thus there exists an initial ordinal $\kappa$ such that no injection $\kappa \to A$ exists, as required. [/step]