# Does local strict contractibility imply ANR?

Say that a space (= compact metrizable space) $X$ is locally strictly contractible if, for every $p\in X$ and neighborhood $U$ of $p$, there is a neighborhood $V$ of $p$ which can be contracted to $p$ within $U$ relative to $p$, i.e. by a homotopy leaving $p$ fixed. (This may be bad terminology, and I welcome alternatives.)

Every ANR is locally strictly contractible. What about the converse? Borsuk's example of a locally contractible space which is not an ANR is not locally strictly contractible.

Incidentally, "locally contractible" is defined in different ways in various references. What is the currently accepted definition?

• Bruce: beware that 'strictly contractible' does have another meaning in the literature, not unrelated to yours: sciencedirect.com/science/article/pii/S0166864101000086 – Sergey Melikhov Jul 20 '12 at 18:08
• Upon some reflection, Borsuk's example does actually seem to be, hmm, pointed locally contractible. It's an interesting question though if there exists a locally contractible compactum that is not pointed locally contractible. It would have to be infinite-dimensional. As for Borsuk's example, it is not locally equi-connected in the sense of Fox and Dugundgi (or equivalently, not uniformly locally contractible in the sense of Isbell). A long-standing open problem which has considerable literature is whether local equi-connectedness is equivalent to the ANR property for compacta. – Sergey Melikhov Jul 20 '12 at 23:06
• I guess I still don't see how to contract Borsuk's example to, say, the point with $x_1=0$ and $x_n=1/2n$ for $n>1$. It seems that any contraction of a neighborhood has to send this point via a point with $n$'th coordinate 0 or $1/n$ for some $n>1$. – Bruce Blackadar Jul 21 '12 at 14:06
• The problem is that to contract a neighborhood, you have to be able to work in a coordinate $m$ for which the neighborhood doesn't hit both ends of $X_m$, so points can be moved around one end. But for any neighborhood of the point $x$, there are only finitely many such coordinates to work with, so the end you move around can't be passed off to infinity. (Hope this cryptic description makes sense.) – Bruce Blackadar Jul 21 '12 at 16:44
• Bruce, I now see that you were right. The motion I've described is of the order of $1/i$ but it affects points that about $(1/m)$-close to $x$. Here $m$ could be large, so for the homotopy to be continuous, $x$ has to move as well. Sorry for taking your time, and thank you for explaining Borsuk's example, which as you see I didn't quite understand. – Sergey Melikhov Jul 22 '12 at 21:37