I was just watching a lecture showing that the cantor set has measure zero, but is still uncountable.
The uncountability argument amounted to taking a number in $[0,1]$ and representing by its ternary expansion, and then arguing that the cantor set is effectively removing all the numbers with a $1$ somewhere in their ternary expansion. From here, we're only left with numbers that have $0$ or $2$, which is in bijection with sequences of $\{0,1\}$ and is thus uncountable.
But it just dawned on me that, even though the cantor set is uncountable, it having measure zero would imply that many "more" numbers than the cantor set have a $1$ in their ternary expansion than numbers that do not.
This seems really surprising to me, but I don't think it should be surprising to me; I think there's some subtlety that I'm misunderstanding. So I guess my real question is; if we were to do a similar construction, but we removed numbers with a $0$ somewhere in their ternary expansion, would we still get a set of measure zero?
Looking at the construction, I think we would still get a set of measure zero; at the $n^{th}$ of construction, we're still reducing the set to being a length of $(\frac{2}{3})^n$, which diminishes to zero.