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In part of Kenneth Falconer's proof of the countable stability of Hausdorff dimension on p. 49, sect 3.2 of Fractal Geometry, I understand him to say that $$\dim_H \bigcup_{i=1}^{\infty}F_i\leq \sup_{1\leq i\leq\infty}\{\dim_H F_i\}\;,$$ because when $s>\dim_H F_i$ for all $i$, ${\cal H}^s(F_i)=0$, and thus $${\cal H}^s(\bigcup_{i=1}^{\infty} F_i) \leq \sum_{i=1}^{\infty}{\cal H}^s(F_i) = 0\;.$$ Here $\dim_H$ is Hausdoff dimension, and ${\cal H}^s$ is $s$-dimensional Hausdorff measure.

I understand everything after $s>\dim_H F_i$ above, but I'm confused about why ${\cal H}^s(\bigcup_{i=1}^{\infty} F_i) \leq \sum_{i=1}^{\infty}{\cal H}^s(F_i) = 0$ implies $\dim_H \bigcup_{i=1}^{\infty}F_i\leq \sup_{1\leq i\leq\infty}\{\dim_H F_i\}$.

I understand that Hausdorff dimension of a set $G$ is the Hausdorff measure for which ${\cal H}^s(G)$ is finite, such that for $s>\dim_H G$, the Hausdorff measure is $0$, and for $s<\dim_H G$, the Hausdorff measure is infinite. I don't see why the fact that ${\cal H}^s(\bigcup_{i=1}^{\infty} F_i) \leq 0$ for an $s$ that is larger than the dimension implies the first inequality above, which concerns a $\sup$ for Hausdorff dimensions that are possibly greater than $0$. I'm sure there must be something obvious that I'm not seeing. I've been thinking about it for a week and I'm still confused.

This answer gives a detailed proof of the part I already understand, but leaves my question unanswered.

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    $\begingroup$ Looking quickly at this (I'm in the middle of something else), I think one approach is to assume, for a later contradiction, that $\sup_{1\leq i\leq\infty}\{\dim_H F_i\} < \dim_H \bigcup_{i=1}^{\infty}F_i.$ As a consequence of this strict inequality, it follows that you can sandwich a real number $s$ strictly between these two quantities, namely $\sup_{1\leq i\leq\infty}\{\dim_H F_i\} < s < \dim_H \bigcup_{i=1}^{\infty}F_i,$ and I believe it's not difficult to get a contradiction from this. For instance, note that $\dim_H F_i < s$ for each $i,$ so $\dots$ $\endgroup$ Sep 15, 2020 at 16:24
  • $\begingroup$ Thanks @DaveL.Renfro. Thinking about that now. $\endgroup$
    – Mars
    Sep 16, 2020 at 12:55

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