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I have a space $A$ which is homeomorphic to the open $n$-ball $B_n$. I'm trying to build a CW-complex with it, so I want a continuous function from the closed ball $\overline{B}_n$ to the closure $\overline{A}$. The spaces I'm working with are relatively nice, they are embedded in the unit cube $(0,1)^m$ for $m\geq n$. I know that this alone is not enough, as $\sin(\frac{1}{x})$ can be shifted and shrunk into the unit square.

On the other hand, I know that if there is a uniformly continuous map from $B_n$ to $A$, then it will preserve Cauchy sequences, so it can be extended to a continuous map on the closures. Furthermore, if a continuous map exists on the closures, the restrictions to the interior has to preserve Cauchy sequences.

So my general question is, what kind of properties could I look for in $A$ that would guarantee that a uniformly continuous map exists, or at least one that preserves Cauchy sequences.

More specifically, I feel like it should work as long as $A$ has finite volume (surface volume? measure? whatever the correct generalization of arc length and surface area would be). Are there any theorems that might say something along those lines?

Edit: The open ball $B_n$ is uniformly homeomorphic to the open hypercube $(0,1)^n$. Let $F:(0,1)^n\rightarrow A$ be a homeomorphism with the property that there exists $K>0$ such that for any $i\in [n]$ and $x\in (0,1)^{n-1}$, the arc length of $f_x(t) = F(x_1,x_2,\ldots, x_{i-1},t,x_i,\ldots,x_{n-1})$ is less than $K$. Does this imply that $A$ is uniformly homeomorphic to $B_n$?

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In three dimensions, merely having finite volume and surface area isn't enough. You could start with something like the helix $(\cos t, \sin t, 1/t)$ for $1 \leq t < \infty$ and then thicken it to get a topological ball, but if the amount of thickening dies off fast enough as $t \to \infty$ then I think both volume and surface area can be finite.

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  • $\begingroup$ Hmm, that's a nice counterexample. What if we include a uniform bound on being connected locally, something like: there exists $\epsilon>0$ such that for all $x\in A$, the intersection of $A$ and the $\epsilon$-ball around $x$ is connected. It seems like that condition might even imply the finite volume. $\endgroup$
    – David Collins
    Commented Jun 24, 2013 at 21:59
  • $\begingroup$ No, I don't think that's quite right. I think I mean there exists $\epsilon>0$ such that for every $\delta$ with $0<\delta<\epsilon$ and for every $x\in A$, the intersection of $A$ with the $\delta$ ball around $x$ is connected. $\endgroup$
    – David Collins
    Commented Jun 25, 2013 at 15:58
  • $\begingroup$ Nope, I think I'm missing the point. It's not that it gets close together, it's that some part of it is infinitely long. $\endgroup$
    – David Collins
    Commented Jun 25, 2013 at 16:56

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