[proofplan] The proof is the direct normal-measure form of Fodor's lemma. The selected column assignment $\alpha \mapsto \beta(\alpha)$ is regressive on the $U$-measure-one set $S$, because each selected column lies strictly below its row index. Applying [Fodor's lemma for normal measures](/theorems/7414) gives a single ordinal $\beta_0$ on which the selection map is constant on a $U$-measure-one subset of $S$. Translating this constant-value set back into the language of columns gives the desired concentration on one column. [/proofplan] [step:Recognize the selected column map as a regressive function] The function \begin{align*} \beta: S \to \kappa \end{align*} is given in the statement, and its defining hypothesis is that $\beta(\alpha) < \alpha$ for every $\alpha \in S$. Thus $\beta$ is a regressive function on the $U$-measure-one set $S$ in the sense relevant to normal measures. If one's convention for regressive functions excludes the ordinal $0$ from the domain, there is no additional issue: the inequality $\beta(\alpha) < \alpha$ cannot hold when $\alpha = 0$, since there is no ordinal below $0$. Hence $0 \notin S$ under the stated hypotheses. [guided] We want to use normality of $U$, and the standard way normality is used is through regressive functions. Here the selected column assignment is not an extra object to construct: it is exactly the function \begin{align*} \beta: S \to \kappa \end{align*} from the statement. A function on a set of ordinals is regressive when its value at each row lies below that row. The theorem assumes precisely this: \begin{align*} \beta(\alpha) < \alpha \end{align*} for every $\alpha \in S$. Therefore the map $\beta$ is regressive on $S$. There is a small endpoint convention that sometimes appears in statements of Fodor's lemma: some authors define regressive functions only on subsets of $\kappa \setminus \{0\}$. This convention does not change the argument here. If $\alpha = 0$, then the inequality $\beta(0) < 0$ is impossible, because no ordinal is strictly below $0$. Since the hypothesis says $\beta(\alpha) < \alpha$ for every $\alpha \in S$, it follows that $0 \notin S$. Thus the domain already lies inside the part of $\kappa$ on which the regressive-function hypothesis makes sense. [/guided] [/step] [step:Apply normality through Fodor's lemma] Since $S \in U$ and $\beta: S \to \kappa$ is regressive, Fodor's lemma for normal measures applies to $\beta$ on $S$ (citing a result not yet verified in the wiki: Fodor's Lemma for Normal Measures). Therefore there exists an ordinal $\beta_0 < \kappa$ such that \begin{align*} \beta^{-1}(\{\beta_0\}) \in U. \end{align*} By the definition of preimage, \begin{align*} \beta^{-1}(\{\beta_0\}) = \{\alpha \in S : \beta(\alpha) = \beta_0\}. \end{align*} Hence \begin{align*} \{\alpha \in S : \beta(\alpha) = \beta_0\} \in U. \end{align*} [/step] [step:Translate the constant-value set into column concentration] The set \begin{align*} \{\alpha \in S : \beta(\alpha) = \beta_0\} \end{align*} is exactly the set of rows in $S$ whose selected column is the fixed column $\beta_0$. Since this set belongs to the normal measure $U$, the row-by-row selection concentrates on the single column $\beta_0$ on a $U$-measure-one set of rows. This is the asserted conclusion. [/step]