MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have encountered the following property. Can anybody tell me if it already exists in literature and/or is equivalent/similar to other well-known properties?

Property: $(X,d)$ metric space. For any open ball $B\subseteq X$ and for any distinct $x,y\in B$, there exist two disjoint open balls $B_1\ni x$ and $B_2\ni y$ and two open continuous and injective functions $f_i:B\rightarrow X$ such that $f_i(B)\subseteq B_i$.

Well.. it's similar to contractibility, but seems to be weaker - it's some density condition..

share|cite|improve this question
Assume that your space is discrete. Then $X$ itself is an open ball and so are '$\{x\}$' and '$\{y\}$'. But injective functions mapping $X$ into the smaller open balls cannot exists. Am I missing something? – Abel Stolz May 28 '11 at 10:13
Perhaps I missed that your statement was entitled 'Property' and not 'Proposition'... What kind of examples possessing the above property do you have in mind? – Abel Stolz May 28 '11 at 10:25
Of course. That property is not always true and indeed it describes the opposite situation for a metric space to be discrete.. For instance any Banach space, but also $\mathbb Q$ with the standard metric – Valerio Capraro May 28 '11 at 10:45
I don't see how this property relates to contractibility. To me at least the stronger property, with one but arbitrarily small ball instead of two disjoint balls, is strongly reminiscent of self-similar sets and fractals. – Sergey Melikhov May 28 '11 at 16:48
@ Bill, it seems to me they are different: let $X$ be the closure in $\mathbb R$ of $\bigcup_{n=0}^\infty[\frac{1}{2^{2n-1}},\frac{1}{2^{2n}}]$. It seems to me that $X$ verifies "my" property, but not yours. Am I right? – Valerio Capraro May 29 '11 at 2:12
up vote 1 down vote accepted

An open continuous injective map can also be described as a homeomorphism with an open subset. The stated property is obviously equivalent to: every open set in every open ball $B$ in $X$ contains an open set that is homeomorphic with $B$. If the metric is bounded, $B$ can be replaced by $X$ without loss of generality (and so the property becomes purely topological).

This property is a kind of local self-similarity (or just self-similarity in the bounded case).

The property does not hold for contractible nor for locally contractible spaces, for instance it fails already for the closed unit interval. It fails even "locally" for the triod, in the sense that the vertex of the triod has no neighborhood satisfying the property.

share|cite|improve this answer
From Wikipedia: a compact topological space $X$ is said to be self-similar if there are non-surjective homeomorphisms $f_1,..f_n$ such that $X=\bigcup f_i(X)$. mmm yes, probably if the space is compact, these two properties are exactly the same.. By the way, why does the definition of self-similarity require compactness? – Valerio Capraro May 29 '11 at 21:34
No, they are not exactly the same. There are many versions of this definition in the literature; usually the $f_i$ are required to be metric contractions or even affine similitudes with ratio $<1$. I don't know if your exact version has appeared anywhere. – Sergey Melikhov May 30 '11 at 11:43
I can't see at the moment an example of compact metric space satisfying "my" condition, but not the wikipedia definition of self-similarity.. nor the converse.. – Valerio Capraro May 30 '11 at 15:16
The closed unit interval $[0,1]$ satisfies the wikipedia definition, but not your condition, because for $x=1/3$ and $y=2/3$ (or for any other pair of distinct $x$ and $y$) there exists no open set in $[0,1]$ containing $x$, disjoint from $y$, and homeomorphic to $[0,1]$. Note that $[0,1]$ is the open ball (in itself) of radius $2$ about the point $1/2$. – Sergey Melikhov May 30 '11 at 16:44
I'm sorry but I don't think it is a counterexample, because I don't want an homeomorphism, but just a topological embedding (i.e. a continuous, open and injective map) and so it suffices to take a closed subinterval of two disjoint open balls centered in $x$ and $y$. – Valerio Capraro May 30 '11 at 21:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.