Typo fixes.
All checks were successful
Compile Project / Compile (push) Successful in 18s

This commit is contained in:
Bokuan Li
2026-03-17 15:30:41 -04:00
parent 16e6beb117
commit dd49a42b4a
2 changed files with 3 additions and 3 deletions

View File

@@ -99,9 +99,9 @@
k(N, x) = \min\bracs{n \in C(N, x) \bigg | d(f(x), y_n) = \min_{m \in C(N, x)}d(f(x), y_m)}
\]
then for any $k \in \natp$,
then for any $k \in \natp$, $\bracs{x \in X|k(n, x) \le k}$ is equal to
\[
\bracs{x \in X|k(n, x) \le k} = \bigcup_{j = 1}^k\bracs{x \in X \bigg |y_j \in C(n, x), d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)}
\bigcup_{j = 1}^k\bracs{x \in X \bigg |y_j \in C(n, x), d(f(x), y_j) = \min_{m \in C(N, x)}d(f(x), y_m)}
\]
For each $1 \le m \le N$, $y \mapsto d(y, y_m)$ is continuous. Thus $x \mapsto d(f(x), y_m)$ and $x \mapsto \min_{m \in C(N, x)}d(f(x), y_m)$ are $(\cm, \cb_\real)$-measurable by \autoref{proposition:limit-measurable} and assumption (c). By \autoref{proposition:metric-measurables},