# A density condition for metric spaces

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..

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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

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).
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
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