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Let $M$ be a metric space and $S\subset M$ a dense subset. For vague reasons (below), it seems to me that the upper box-counting dimension of $S$ should be equal to that of $M$, but I don't quite see the argument, and I am skeptical of my reasons anyway.

Is the upper box-counting dimension of $S$ equal to that of $M$? If so, what's the argument? If not, what's a counterexample?

Definitions and thoughts:

The upper box-counting dimension of a metric space $M$ is defined as follows: for $\varepsilon > 0$, let $N(\varepsilon)$ represent the minimal cardinality of a cover of $M$ by $\varepsilon$-balls. Then the upper box-counting dimension is

$$\limsup_{\varepsilon\to 0} \frac{\log N(\varepsilon)}{\log 1/\varepsilon}$$

Now if $S$ is dense in $M$, a cover of $S$ by $\varepsilon$-balls is "almost" a cover of $M$; the complement of the cover is a closed set with no interior. The number of extra $\varepsilon$-balls needed to cover the rest of $M$ should thus be "small". This is the intuition behind my question.

But I can't think of a way to make this argument at all quantitative because all the facts I have about the complement of the cover are topological - it is closed with empty interior, i.e. it is closed and nowhere dense - but not metric. In $\mathbb{R}$ we can easily have closed, nowhere dense sets that are not "small" in other ways - e.g. fat Cantor sets are closed and nowhere dense but of positive measure. So although this doesn't directly speak to the case at hand, it does suggest that the intuition mentioned above is not reliable.

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  • $\begingroup$ If you can show that your definition doesn't depend on if your balls are open or closed, you've basically solved the question, because every (finite) cover of $S$ by closed balls cover $M$. $\endgroup$
    – Tryss
    May 2, 2015 at 15:38
  • $\begingroup$ @Tryss - that's more or less exactly the question, isn't it? $\endgroup$ May 2, 2015 at 16:02
  • $\begingroup$ @BenBlumSmith : no, it's a more general question, and easier to answer. $\endgroup$
    – Tryss
    May 2, 2015 at 16:46
  • $\begingroup$ @Tryss - Ah! I think you're suggesting I note that open $\varepsilon$-balls contain closed $\varepsilon/2$-balls, etc. $\endgroup$ May 2, 2015 at 16:56

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Let's prove that the definition doesn't depend if the ball is open or closed.

The cover number for closed balls, noted $N_c$ verify

$$N(\epsilon(1+\epsilon) ) \leq N_c(\epsilon) \leq N(\epsilon)$$

Hence

$$\frac{\log N(\epsilon(1+\epsilon) ) } {\log (\frac{1}{\epsilon}) } \leq \frac{\log N_C(\epsilon ) } {\log (\frac{1}{\epsilon}) } \leq \frac{\log N(\epsilon ) } {\log (\frac{1}{\epsilon}) }$$

$$\frac{\log (\frac{1}{\epsilon(1+\epsilon)})}{\log (\frac{1}{\epsilon})} \frac{\log N(\epsilon(1+\epsilon) ) } {\log (\frac{1}{\epsilon(1+\epsilon)}) } \leq \frac{\log N_C(\epsilon ) } {\log (\frac{1}{\epsilon}) } \leq \frac{\log N(\epsilon ) } {\log (\frac{1}{\epsilon}) }$$

And as $\frac{\log (\frac{1}{\epsilon(1+\epsilon)})}{\log (\frac{1}{\epsilon})} \to 1$, we can see that

$$\lim_{\epsilon \to 0} \frac{\log N_C(\epsilon ) } {\log (\frac{1}{\epsilon}) } = \lim_{\epsilon \to 0} \frac{\log N(\epsilon ) } {\log (\frac{1}{\epsilon}) }$$

Hence the definition doesn't depend if the balls are closed or open.

Then the initial problem is solved trivially by using the definition with closed balls

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  • $\begingroup$ I buy this in essence, although there's something to be careful of since I am defining the dimension in terms of $\limsup$ and not assuming that a limit exists... $\endgroup$ May 4, 2015 at 0:37
  • $\begingroup$ It shouldn't change anything to the demonstration, let me check $\endgroup$
    – Tryss
    May 4, 2015 at 0:53

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