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Call a topology "locally self-similar" if it has a basis in which each open set is homeomorphic to the entire space. What topologies have this property?

So far, I have the following list:

  • Any set with the indiscrete topology (the whole space is the unique neighborhood of any point).
  • The real numbers.
  • The rational numbers (as a subspace of the real numbers).
  • Probably the Cantor set or something similar (I'm not sure whether the endpoints look locally like the other points).
  • Probably the Sierpinski carpet and lots of similar spaces.
  • Probably the irrational numbers.
  • Any finite product of spaces with this property.

Anything else? Is it possible to classify these spaces in any interesting way?

Hew Wolff
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  • There is also the right (or left) order topology on the real numbers, which is not only locally self-similar but self-similar (@BrianMScott's "stronger property"). The same construction seems to work for the rational and irrational numbers: make a new topology whose open sets are the existing open sets intersected with all $(x, \infty)$ for all real $x$. – Hew Wolff Sep 21 '12 at 21:07
  • Amazing question. I am sure there is some sufficient condition in terms of action of a semigroup (I can see how it arises in examples with reals, rationals, totally disconnected spaces, as well as some non-Hausdorff (albeit T₀) spaces not mentioned by original poster). Will think on it further. – Incnis Mrsi Nov 09 '14 at 18:57
  • Great, @IncnisMrsi, tell me what you find. – Hew Wolff Nov 14 '14 at 17:40

1 Answers1

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The Cantor set is indeed an example: it’s homeomorphic to $\{0,1\}^\omega$, where $\{0,1\}$ has the discrete topology, and so is every basic open set in this product. Indeed, if $X$ is any discrete space, and $\kappa$ is any infinite cardinal, $X^\kappa$ has the property: every member of the obvious base for the product topology is clearly homeomorphic to $X^\kappa$. The Sierpiński carpet is a Cantor set.

The Cantor set is almost an example of a stronger property: it’s almost a space in which every non-empty open set is homeomorphic to the whole space. It actually has two kinds of non-empty open subset, compact ones, which are homeomorphic to the Cantor set, and non-compact ones, which are homeomorphic to the Cantor set minus a point and to the discrete union of $\omega$ copies of the Cantor set.

The irrationals are an example of a space with the stronger property: by an old result of Alexandroff and Uryson they are the unique topologically complete, separable, $0$-dimensional metric space that contains no non-empty compact open set, and all of these properties are inherited by non-empty open subsets. More generally, if $X$ is any infinite discrete space, $X^\omega$ is a metrizable space of weight $|X|$ with the stronger property.

If $\kappa$ is any infinite cardinal, $\{\kappa\setminus\alpha:\alpha<\kappa\}\cup\{\varnothing\}$ is a $T_0$ topology on $\kappa$ in which all non-empty open sets are homeomorphic.

If $\lambda\le\operatorname{cf}\kappa$ is also an infinite cardinal, $\{U\subseteq\kappa:|\kappa\setminus U|<\lambda\}\cup\{\varnothing\}$ is a $T_1$ topology on $\kappa$ with the desired property; when $\lambda=\omega$ it’s simply the cofinite topology.

Brian M. Scott
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  • Fascinating. Let's call the stronger property "self-similar" (every nonempty open set is homeomorphic to the whole space). Are you saying that the irrationals is an example of the $X^\omega$ construction for self-similar spaces? – Hew Wolff Sep 16 '12 at 23:53
  • @Hew: Yes: the irrationals are homeomorphic to $\omega^\omega$ with the product topology. This can be proved using continued fraction expansions of the irrationals or directly, as in [these notes](http://arxiv.org/pdf/math/9401202). – Brian M. Scott Sep 16 '12 at 23:57
  • I think that if $X$ is a discrete space and $\kappa$ is finite, then $X^\kappa$ is indeed discrete, but it does not have the LSS property. (Points are open but not homeomorphic to the whole space.) $\kappa$ must be infinite to make that work. – Hew Wolff Sep 16 '12 at 23:58
  • @Hew: Yes, that was a silly mental hiccup on my part; it’s fixed now. – Brian M. Scott Sep 17 '12 at 00:01
  • You describe the Sierpinski carpet as a Cantor set, but my reading suggests that it is not homeomorphic to the Cantor set. For example, it appears that the Sierpinski carpet contains any graph as a subspace, and it's hard to imagine those graphs fitting into the Cantor set. If that's the case, I wonder whether the Sierpinski carpet is LSS or even SS. – Hew Wolff Sep 17 '12 at 16:59