= Solution
Take the middle-thirds <Cantor set> $C$. Its $n$th stage is the union of $2^n$ closed intervals of length $3^{-n}$, and contains $C$. Their total length is $(2/3)^n\to0$; slightly enlarging them to open intervals gives covers of arbitrarily small total length. Therefore \b[$C$ is null].
For uncountability map every <binary sequence> $(\varepsilon_j)_{j\geq1}$ to $\sum_{j\geq1}2\varepsilon_j3^{-j}\in C$. If two sequences first differ at index $k$, their image difference in absolute value is at least $2\cdot3^{-k}-\sum_{j>k}2\cdot3^{-j}=3^{-k}>0$. Thus the map is injective. The set of <binary sequences> is <uncountable> by <Cantor's diagonal argument>, so \b[$C$ is <uncountable>].
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