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Keywords:
Hilbert cube; absorbing system; $F_\sigma$; $F_{\sigma \delta}$; capset; Peano continuum; hyperspace; hyperspace of subcontinua; covering dimension; cohomological dimension
Hilbert cube; absorbing system; $F_\sigma$; $F_{\sigma \delta}$; capset; Peano continuum; hyperspace; hyperspace of subcontinua; covering dimension; cohomological dimension
Summary:
Let $\Cal D$ denote a true dimension function, i.e., a dimension function such that $\Cal D(\Bbb R^n) = n$ for all $n$. For a space $X$, we denote the hyperspace consisting of all compact connected, non-empty subsets by $C(X)$. If $X$ is a countable infinite product of non-degenerate Peano continua, then the sequence $(\Cal D_{\geq n}(C(X)))_{n=2}^\infty$ is $F_\sigma$-absorbing in $C(X)$. As a consequence, there is a homeomorphism $h: C(X)\rightarrow Q^\infty$ such that for all $n$, $h[\{A \in C(X) : \Cal D(A) \geq n+1\}] = B^n \times Q \times Q \times \dots $, where $B$ denotes the pseudo boundary of the Hilbert cube $Q$. It follows that if $X$ is a countable infinite product of non-degenerate Peano continua then $\Cal D_{\geq n}(C(X))$ is an $F_\sigma$-absorber (capset) for $C(X)$, for every $n \geq 2$. Let $\operatorname{dim}$ denote covering dimension. It is known that there is an example of an everywhere infinite dimensional Peano continuum $X$ that contains arbitrary large $n$-cubes, such that for every $k \in \Bbb N$, the sequence $(\operatorname{dim}_{\geq n}(C(X^k)))_{n=2}^\infty$ is not $F_\sigma$-absorbing in $C(X^k)$. So our result is in some sense the best possible.
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