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To be specific, my question is as follows:

Question: Let $X$ be an inverse limit of compact metric spaces $(X_i, d_i)$, then does it hold

$\dim(X, d) \leq \sup_i \{\dim (X_i, d_i)\}$ for some compatible metric $d$ on $X$?

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What is your notion of inverse limit of metric spaces? Do you mean this one: mathoverflow.net/questions/15948/… ? –  Misha May 19 '13 at 2:37
    
Yes, and here metric dimension means the low box dimension. –  Bingbing Liang May 19 '13 at 2:41
    
Which spaces are inverse limits of finite metric spaces? –  Gerald Edgar Feb 10 at 18:33

1 Answer 1

I really should avoid answering questions late at night. My original answer is muddled enough to not work. But here is what it should have been:

Let $X_0 = \{0,1\}$ and $X_i = X_{i-1} \times X_0$. Give each $X_i$ the $2-$adic ultrametric. That is the distance between two sequences of $0'$s and $1'$s is $2^{-n}$ where $n$ is the number of share initial symbols the two sequences share. The projections maps are given by truncation. The inverse limit is now $\mathbb{Z}_2$ which has box counting dimension equal to one.

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This construction does not seem to be an inverse limit of metric spaces in Petrunin's sense (mathoverflow.net/questions/15948/…). –  Misha May 19 '13 at 4:29
    
Even it fails in the general setting, how about making X_i to be the compact metric (abelian) group? –  Bingbing Liang May 19 '13 at 4:58
    
@Misha thank you for making me think a bit longer about this. @Bingbing Liang yes, the idea is to do that and then take the limit. Z_2 is just the easiest example of this to write down. –  BSteinhurst May 19 '13 at 5:42
    
Good, now it works. –  Misha May 19 '13 at 13:20
    
In general, the low box dimension is dependent on the choice of metric. @BSteinhurst Is it clear that each compatible metric on $\mathbb{Z}_2$ induces a nonzero low box dimension? –  Bingbing Liang May 21 '13 at 2:50

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