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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$\begingroup$ What is your notion of inverse limit of metric spaces? Do you mean this one: mathoverflow.net/questions/15948/… ? $\endgroup$– MishaCommented May 19, 2013 at 2:37
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$\begingroup$ Yes, and here metric dimension means the low box dimension. $\endgroup$– Bingbing LiangCommented May 19, 2013 at 2:41
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$\begingroup$ Which spaces are inverse limits of finite metric spaces? $\endgroup$– Gerald EdgarCommented Feb 10, 2014 at 18:33
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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.
answered May 19, 2013 at 3:49
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$\begingroup$ This construction does not seem to be an inverse limit of metric spaces in Petrunin's sense (mathoverflow.net/questions/15948/…). $\endgroup$– MishaCommented May 19, 2013 at 4:29
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$\begingroup$ Even it fails in the general setting, how about making X_i to be the compact metric (abelian) group? $\endgroup$ Commented May 19, 2013 at 4:58
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$\begingroup$ @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. $\endgroup$ Commented May 19, 2013 at 5:42
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$\begingroup$ 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? $\endgroup$ Commented May 21, 2013 at 2:50