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Source. Theorem 1, p. 712, proof p. 712, of R. D. Anderson and J. E. Keisler, An example in dimension theory, Proc. Amer. Math. Soc. 18 (1967), no. 4, 709--713, DOI 10.1090/S0002-9939-1967-0215288-0, the edition named on the source card.

Statement

Setting (p. 709). $\omega$ is the set of positive integers, $E^n$ is Euclidean $n$-space, $\dim$ is the (inductive) topological dimension of Hurewicz and Wallman, and $K^s$ is the product of $s$ copies of $K$.

Theorem 1 (p. 712, quoted). "Let $n$, $s\in\omega$. There exists $K\subset E^n$ such that $\dim K=\dim K^s=n-1$."

Here the set may depend on both $n$ and $s$; the paper's Theorem 2 removes the dependence on $s$ and adds the countable power.

Read depth. Claims checked: the statement and the setting were read clause by clause on the page images of pp. 709 and 712. The proof was read in outline only; its steps and Lemmas 1--3 were not checked. Nothing here is independently reviewed.

Proof pointer

Page 712, written here in outline. Lemma 3 (p. 711) supplies countably many $(ns-n)$-spheres $S_i$ in $E^{ns}=(E^n)^s$ such that any $T\subset E^{ns}$ missing all of them has $\dim T\le n-1$. The proof well-orders the nondegenerate continua of $E^n$ so that each has fewer than $\mathfrak c$ predecessors and builds $K$ by transfinite induction: whenever a continuum does not yet meet the set built so far, one of its points is added, chosen so that no $s$-letter word over the enlarged set lies on any $S_i$. This is possible because each continuum has $\mathfrak c$ points while fewer than $\mathfrak c$ of them are excluded. Then $K^s$ misses every $S_i$, so $\dim K^s\le n-1$ and hence $\dim K\le n-1$; and $K$ meets every nondegenerate continuum, so Lemma 1 (p. 710) gives $\dim K\ge n-1$, and therefore $\dim K^s\ge n-1$.

Dependencies

Lemmas 1 and 3 of the same paper (pp. 710--711), and Lemma 2 (pp. 710--711), which the paper uses in a weakened form "without explicit proof here" (p. 710).

Bears on

• Problem 909: the problem asks, for $n\ge2$, for a space $S$ of dimension $n$ with $S^2$ also of dimension $n$. Theorem 1 with $s=2$, applied in $E^{n+1}$, gives a set $K\subset E^{n+1}$ with $\dim K=\dim K^2=n$, for every $n\ge1$, with $\dim$ the inductive dimension of Hurewicz and Wallman. The paper does not mention the problem.