3
$\begingroup$

Given a metric space $X$, denote by $\mathrm{Haus}\,X$ the space of all compact subsets in $X$ equipped with Hausdorff metric. Further $X$ will be identified with a subset of $\mathrm{Haus}\,X$ --- a point $x\in X$ corresponds to the one-point set $\{x\}\subset X$.

A metric space $X$ will be called nice if there is a short (= 1-lipschitz) retract $\mathrm{Haus}\,X\to X$. In other words, if for any compact set $A\subset X$ there is a point $p_A\in X$ such that $p_{\{x\}}=x$ and $$|p_A-p_B|_X\leqslant |A-B|_{\mathrm{Haus}\,X}.$$ (Note that we do not assume that $p_A\in A$.)

For example, any injective space is nice. Also, discrete spaces (all distances = 1) are nice. On the other hand, Euclidean plane is not nice.

Is there an example of a nice compact geodesic space which is not injective?

Improve this question
$\endgroup$
6
  • $\begingroup$ The Lipschitz niceness condition can also be written as: for any $X,Y$, there are $x\in X,\,y\in Y$ with $d(x,y)\ge d(f(X),f(Y))$ and $d(x,y)$ equal to either $d(x,Y)$ or $d(X,y)$. $\endgroup$
    – user44143
    Commented Nov 23, 2022 at 15:27
  • $\begingroup$ @MattF. No, I do not assume $r(A)\in A$. $r(A)$ might be any point such that $r(\{x\})=x$ and $|r(A)-r(B)|\leqslant|A-B|_H$. $\endgroup$
    – Anton Petrunin
    Commented Nov 23, 2022 at 15:32
  • $\begingroup$ This is just another way of saying that the Hausdorff distance between $X$ and $Y$ is greater than that between $f(X)$ and $f(Y)$ — so the $x\in X,y\in Y$ there just comes from the definition of Hausdorff distance. Without that clause the condition I wrote would be trivial; as it stands it is equivalent to the condition in the post. $\endgroup$
    – user44143
    Commented Nov 23, 2022 at 15:43
  • 1
    $\begingroup$ If $(X,d)$ is a metric space and $\epsilon \in (0,1)$, then $(X, d^\epsilon)$ is also a metric space (called a ``snowflake'' of $(X,d)$). It seems to me that if the former space is nice, then so is the latter, trivially. The snowflake will not be injective. For a concrete example, take $X=[0,1]$ and $\epsilon=\frac{1}{2}$. Does this work or have I made a mistake? $\endgroup$
    – anon
    Commented Nov 27, 2022 at 1:50
  • $\begingroup$ @anon I asked for geodesic space, and snowflake is not geodesic. But it is a nice observation. $\endgroup$
    – Anton Petrunin
    Commented Nov 28, 2022 at 10:33

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .