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Garrett Ervin
Garrett Ervin
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As was conjectured by Adam in his answer, every finite $n > 2$ has the property you're looking for. Namely, there exists a topological space homeomorphic to its $n$th power and none of its $k$th powers, $1<k<n$.

The relevant paper is "Homeomorphisms between finite powers of topological spaces" by Orsatti and Rodino. It can be found here. From the abstract:

"Let $\lambda$ be an infinite cardinal number. It is proved that, for each positive integer $r$, there exists a compact connected homogeneous topological space $X$ of weight $\lambda$ such that $X^n$ is homeomorphic to $X^m$ iff $n \equiv m$ (mod $r$)."

From glancing at the paper, the authors adapt a big theorem of Corner, that every countable reduced torsion-free ring is the endomorphism ring of some torsion-free Abelian group, to prove their result. (This is the same theorem Corner himself uses to prove the result cited in Adam's answer.) What they actually produce is a compact and connected topological Abelian group $X$ such that $X^n$ is topologically isomorphic $X^m$ iff $n \equiv m$ (mod $r$), and then use the fact that for such groups any homeomorphism ofbetween the underlying spacespaces must actually be a topological isomorphism.

As was conjectured by Adam in his answer, every finite $n > 2$ has the property you're looking for. Namely, there exists a topological space homeomorphic to its $n$th power and none of its $k$th powers, $1<k<n$.

The relevant paper is "Homeomorphisms between finite powers of topological spaces" by Orsatti and Rodino. It can be found here. From the abstract:

"Let $\lambda$ be an infinite cardinal number. It is proved that, for each positive integer $r$, there exists a compact connected homogeneous topological space $X$ of weight $\lambda$ such that $X^n$ is homeomorphic to $X^m$ iff $n \equiv m$ (mod $r$)."

From glancing at the paper, the authors adapt a big theorem of Corner, that every countable reduced torsion-free ring is the endomorphism ring of some torsion-free Abelian group, to prove their result. (This is the same theorem Corner himself uses to prove the result cited in Adam's answer.) What they actually produce is a compact and connected topological Abelian group $X$ such that $X^n$ is topologically isomorphic $X^m$ iff $n \equiv m$ (mod $r$), and then use the fact that for such groups any homeomorphism of the underlying space must actually be a topological isomorphism.

As was conjectured by Adam in his answer, every finite $n > 2$ has the property you're looking for. Namely, there exists a topological space homeomorphic to its $n$th power and none of its $k$th powers, $1<k<n$.

The relevant paper is "Homeomorphisms between finite powers of topological spaces" by Orsatti and Rodino. It can be found here. From the abstract:

"Let $\lambda$ be an infinite cardinal number. It is proved that, for each positive integer $r$, there exists a compact connected homogeneous topological space $X$ of weight $\lambda$ such that $X^n$ is homeomorphic to $X^m$ iff $n \equiv m$ (mod $r$)."

From glancing at the paper, the authors adapt a big theorem of Corner, that every countable reduced torsion-free ring is the endomorphism ring of some torsion-free Abelian group, to prove their result. (This is the same theorem Corner himself uses to prove the result cited in Adam's answer.) What they actually produce is a compact and connected topological Abelian group $X$ such that $X^n$ is topologically isomorphic $X^m$ iff $n \equiv m$ (mod $r$), and then use the fact that for such groups any homeomorphism between the underlying spaces must actually be a topological isomorphism.

Source Link
Garrett Ervin
Garrett Ervin
  • 1.6k
  • 1
  • 13
  • 14

As was conjectured by Adam in his answer, every finite $n > 2$ has the property you're looking for. Namely, there exists a topological space homeomorphic to its $n$th power and none of its $k$th powers, $1<k<n$.

The relevant paper is "Homeomorphisms between finite powers of topological spaces" by Orsatti and Rodino. It can be found here. From the abstract:

"Let $\lambda$ be an infinite cardinal number. It is proved that, for each positive integer $r$, there exists a compact connected homogeneous topological space $X$ of weight $\lambda$ such that $X^n$ is homeomorphic to $X^m$ iff $n \equiv m$ (mod $r$)."

From glancing at the paper, the authors adapt a big theorem of Corner, that every countable reduced torsion-free ring is the endomorphism ring of some torsion-free Abelian group, to prove their result. (This is the same theorem Corner himself uses to prove the result cited in Adam's answer.) What they actually produce is a compact and connected topological Abelian group $X$ such that $X^n$ is topologically isomorphic $X^m$ iff $n \equiv m$ (mod $r$), and then use the fact that for such groups any homeomorphism of the underlying space must actually be a topological isomorphism.