Timeline for Does local strict contractibility imply ANR?
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21 events
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| Jul 23, 2012 at 1:00 | comment | added | Bruce Blackadar | This is a tricky (but worthwhile) example to understand. I'm still not sure I could write out a rigorous proof of what I claimed, but I'm pretty sure it's right. | |
| Jul 22, 2012 at 21:37 | comment | added | Sergey Melikhov | 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. | |
| Jul 22, 2012 at 19:47 | comment | added | Sergey Melikhov | by a motion that is entirely along the segment [1/(m+1),1/m] of the first coordinate, nothing obstructs pulling it further towards $x$, by a linear homotopy. | |
| Jul 22, 2012 at 19:40 | comment | added | Sergey Melikhov | (Starting from the above formula for the belt, I'm dropping the infinite tail of zeros.) All of this belt can be pulled down towards, and then into the subset $\{1/(m+1),1/m\}\times [1/4,1/4]^{i-1}\times\{0\}\times [0,1]^{m-i}$ of the face $[1/(m+1),1/m]\times [0,1]^{i-1}\times\{0\}\times [0,1]^{m-i}$ of the cube with boundary sphere $X_m$ (so this face lies within the sphere). This motion is entirely along the $(i+1)$st coordinate, so it moves each point at most by $1/(i+1)$. As a result, we've taken the belt off the cube's waist, and once we pull a part of it across the bottom of the cube, | |
| Jul 22, 2012 at 19:27 | comment | added | Sergey Melikhov | OK, so Borsuk could be using the $l_2$ metric, and I'm more comfortable using the notation where the infinite product is endowed with the inverse limit metric (as the inverse limit of the finite products). So your $x$ would be (0,1/2,1/2,...) in my notation. A basic neighborhood of this point is of the form $U_i=[0,1/i]\times \prod_{j=i}^0[\frac12-2^{-j},\frac12+2^{-j}]\times [0,1]^\infty$. It meets the sphere $X_m=\partial([1/(m+1),1/m]\times [0,1]^m\times\{0\}^\infty)$, where $m\ge i$, in a subset of the belt $\{1/(m+1),1/m\}\times[1/4,1/4]^i\times [0,1]^{m-i}$. (to be cont'd) | |
| Jul 22, 2012 at 13:18 | comment | added | Bruce Blackadar | This space is a sort of multidimensional version of the infinite comb space (or, more accurately, the infinite ladder), where to contract everything must be moved through a corner point. | |
| Jul 22, 2012 at 2:16 | comment | added | Bruce Blackadar | Our notation seems to be different. I was using Borsuk's notation where the Hilbert cube is the product over all $n$ of $[0,1/n]$. In your notation the point $x$ has first coordinate 0 and all other coordinates 1/2. By the two ends of $X_m$ I meant the end with $x_m=0$ and the end with $x_m=1/m$ (Borsuk's notation). | |
| Jul 22, 2012 at 0:20 | comment | added | Sergey Melikhov | I'm assuming $X_m$ is the boundary of the cube $[1/(m+1),1/m]\times [0,1]^m\times\{0\}^\infty$. Then 'both ends' could mean $\{1/(m+1),1/m\}\times[0,1]^m\times\{0\}^\infty$ or it could mean $[1/(m+1),1/m]\times[0,1]^{m-1}\times\{0,1\}\times\{0\}^\infty$. The latter two sides shouldn't be of concern for any point $x$, and the former one would be of concern for an $x$ that lies close to the cube $Q_m=\{0\}\times [0,1]^m\times\{0\}^\infty$, but far from its boundary. Your $x$ is about as close to $Q_m$ as to its boundary, for each $m$. Every $x$ has only finitely many $Q_m$'s to worry about. | |
| Jul 21, 2012 at 16:44 | comment | added | Bruce Blackadar | 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.) | |
| Jul 21, 2012 at 15:20 | comment | added | Sergey Melikhov | I don't see any problem with this point $x$. Some points $y$ in a neighborhood of $x$ (namely, those with $1/k>y_1>1/(k+1)$) would have to traverse a region $R_k$ consisting of points $z$ with $z_k=0$ (only for a while, not until the end of the homotopy). But for $y$ to be in a neighborhood of $x$, $y_k$ must be bounded above by an $\epsilon_k$, where $\epsilon_k\to 0$ as $k\to\infty$. So the motion of $y$ towards $R_k$ and through $R_k$ becomes smaller and smaller as $y\to x$, and then it doesn't force $x$ to move. | |
| Jul 21, 2012 at 14:06 | comment | added | Bruce Blackadar | 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 $n>1$. | |
| Jul 21, 2012 at 12:47 | comment | added | Bruce Blackadar | OK, I see my mistake in analyzing Borsuk's example. But the concept of local equiconnectedness seems to be the right one for the work I am currently doing, rather than pointed local contractibility. Thanks for pointing me in the right direction. | |
| Jul 21, 2012 at 4:56 | comment | added | Sergey Melikhov | See also ams.org/journals/proc/1983-087-01/S0002-9939-1983-0677252-6 | |
| Jul 20, 2012 at 23:06 | comment | added | Sergey Melikhov | 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. | |
| Jul 20, 2012 at 19:34 | comment | added | Bruce Blackadar | Thanks. Not really being a topologist (although we operator algebra people sometimes like to think our subject includes topology as a special case), I'm not always up on correct topology terminology or the topology literature. What name do you suggest for the property I describe? | |
| Jul 20, 2012 at 18:08 | comment | added | Sergey Melikhov | Bruce: beware that 'strictly contractible' does have another meaning in the literature, not unrelated to yours: sciencedirect.com/science/article/pii/S0166864101000086 | |
| Jul 20, 2012 at 17:57 | comment | added | Sergey Melikhov | David: where does your notion of 'locally P' come from? Is it something originating in algebraic geometry, or just someone's philosophy? I haven't ever seen it used in topology (in the rare event that people need it, they just say 'has P neighborhoods of points'), and I suppose in geometric and general topology there's nearly a century-long tradition of using 'locally P' in the sense of appropriately many neighborhoods (normally one for each quantifier). | |
| Jul 20, 2012 at 2:49 | comment | added | David Roberts♦ | Yes, the contractibility of the open sets as spaces or in their own right, or as subspaces is an important distinction that is not often emphasised. It is a pity there is not a more systematic notation. | |
| Jul 20, 2012 at 1:55 | comment | added | Bruce Blackadar | Borsuk's definition of "locally contractible" is the same as the one I gave of "locally strictly contractible" without the "relative to $p$" part. A stronger definition sometimes used is that $X$ is locally contractible if every point has a neighborhood base of contractible open sets. By this stronger definition even an AR is not necessarily locally contractible. | |
| Jul 19, 2012 at 23:51 | comment | added | David Roberts♦ | A space is generally defined as 'locally P' if there is a neighbourhood basis with each element of the basis having property P. What you have is probably called locally relatively contractible (as is a special case of 'locally relatively P', which refers to a property of the inclusion map of the elements of the basis), or even "$V\to U$ is nullhomotopic as a pointed map". | |
| Jul 19, 2012 at 22:04 | history | asked | Bruce Blackadar | CC BY-SA 3.0 |