---

Statement

Remark (p. 10, unnumbered, quoted). "Let $\mathfrak n<\mathfrak m<\mathfrak c$ be cardinal numbers, and $\{f_\alpha\}$ a family of analytic functions such that for each $z$ the set $\{f_\alpha(z)\}$ consists of at most $\mathfrak n$ distinct values. Then the family has power at most $\mathfrak n$."

The paper introduces it as what the first part of its proof really gives. The cardinal $\mathfrak m$ enters only through the hypothesis that some cardinal lies strictly between $\mathfrak n$ and $\mathfrak c$, that is, $\mathfrak n^+<\mathfrak c$. No hypothesis on the continuum is made.

Proof pointer

P. 10 gives no separate proof; it refers to the first part of the proof of the theorem. Run with $\mathfrak n^+$ functions in place of $\aleph_1$: their pairwise coincidence sets are denumerable, so their union has power at most $\mathfrak n^+\le\mathfrak m<\mathfrak c$, and at a point outside it the $\mathfrak n^+$ values are distinct, more than $\mathfrak n$.

Read depth

Claims checked: the statement was read clause by clause on the page image of p. 10, and the argument it refers to on p. 9 was followed. Nothing here is independently reviewed.

Dependencies

The theorem (p. 9), whose counting argument it reuses.

Source. P. Erdős, An interpolation problem associated with the continuum hypothesis, Michigan Math. J. 11 (1964), 9--10, doi:10.1307/mmj/1028999028, p. 10; the edition read is named on the source card.

Bears on

• Problem 1119: with the problem's $\mathfrak m$ in the role of the remark's $\mathfrak n$, and entire functions being analytic, the remark answers the problem yes for every $\mathfrak m$ with $\mathfrak m^+<\mathfrak c$. It says nothing about the case $\mathfrak m^+=\mathfrak c$.