Let $\mu$ denote that Hausdorff $1$-dimensional measure. Suppose $S\subset\mathbb{B}^m$, where $B^m$ is the unit ball in $\mathbb{R}^m$, $m\geq 3$. I'm not so familiar with the Hausdorff measure, but I understood that up to a multiplicative constant $\mu(S)$ should be equal to the Lebsgue measure in $\mathbb{R}^m$.
I believe that if $S=\{p_1,...,p_n\}$ is a finite collection of points then $\mu(S)=0$. I thought, by similar reasoning to showing that the Lebesgue measure of $\mathbb{Q}$ in $\mathbb{R}$ is zero, that you could show that $\mu(S) = 0$ when $S=\{p_1,p_2,...\}$ is a countable collection of points.
The reason I am confused is because in a talk I went to, I thought it was suggested that countable sets aren't included amongst the sets whose Hausdorff 1-dimensional measure vanishes, since there will be an accumulation point. However, it's possible I misunderstood what the speaker was saying.
So is $\mu(\{p_1,p_2,p_3,...\}) = 0$ for a countable collection of points?
There are a couple of ways of seeing that countable collections of points must have zero one-dimensional Hausdorff measure:
The one-dimensional Hausdorff measure is equal to the one-dimensional Lebesgue measure (perhaps up to multiplication by a constant, depending on how, precisely, the Hausdorff measure is constructed). Every countable set has measure zero with respect to the Lebesgue measure. This last fact needs to be proved, but by the time you are mucking about with Hausdorff measures, you should probably already have proved this. This approach is kind of swatting a gnat with a nuke, but it's... fine.
An important result about the Hausdorff measure is that if $\mathscr{H}^s(E) < \infty$, then $\mathscr{H}^t(E) = 0$ for any $t > s$. That is, if the $s$-dimensional Hausdorff measure of some set is finite, then the measure of that set is zero with respect to the $t$-dimensional Hausdorff measure for any $t > s$. This is an important observation, since the Hausdorff dimension of a space $E$ is defined to be $$\operatorname{dim}_H(E) = \inf\{ s : \mathscr{H}^s(E) = 0\} = \sup\{ s : \mathscr{H}^s(E) = \infty\}. $$
By direct computation: suppose that $E = \{ x_j \}_{j}$ is a countable set. Choose any $\delta > 0$ and any $0 < \varepsilon < \delta$. Observe that $X$ can be covered by the collection of open balls $$ \{ B(x_j, 2^{-j-1} \varepsilon) \}_{j}. $$ But then \begin{align} \mathscr{H}^{1}_{\delta}(E) &= \inf\left\{ \sum_j \operatorname{diam}(U_j)^1 : \text{$\{U_j\}_j$ is a cover of $E$ with $\operatorname{diam}(U_j) < \delta$ for all $j$} \right\} \\ &< \sum_{j} \operatorname{diam}(B(x_j,2^{-j-1}\varepsilon)^1 \\ &= \sum_{j} 2^{-j}\varepsilon \\ &= \varepsilon. \end{align} Hence for any $\varepsilon > 0$, $$ \mathscr{H}^{1}_{\delta}(E) < \varepsilon \implies \mathscr{H}^{1}_{\delta}(E) = 0. $$