Real analysis contradiction I cannot get rid of

Let $G$ be the Cantor set. It is well known that:

1. $G$ is perfect and hence closed.
2. $G$ has the cardinality of the continuum.
3. $G$ has measure zero.
4. For any set $S \subset \mathbb{R}$ (I will not keep writing that we are in $\mathbb{R}$) we have $S \text{ is closed} \Leftrightarrow S^c \text{ is open}$.
5. Any open set $O$ can be written as a --- in fact unique --- countable union of disjoint open intervals.

$G^c$ can thus be written as a countable union of disjoint open intervals. We now imagine this union as being superposed on the real line graphically as follows:

R: <<<----(....)---(..)--(.)---------(...)--->>>

where the (...) represents the open disjoint intervals (of differing size) composing $G^c$, and the --- represents the remaining non-covered real numbers (that is those in $G$). Now we can cover $\mathbb{R}$ in its entirety by "collecting" the --- into disjoint closed intervals. Any one of these closed intervals might of course consist of only a single element. We now obtain:

R: <<<[--](....)[-](..)[](.)[-------](...)[-]>>>

Take the union of these disjoint closed intervals. This must be $(G^c)^c = G$. Now it is not hard to imagine a mapping from the (...)'s to the [...]'s. Just take the next (...) in line for each [...] (and do some trivial fixing at the ends). Therefore we have written $G$ as a countable union of disjoint closed sets. However, $G$ has measure zero and therefore cannot contain any closed sets other than the single element type. Hence $G$ is countable. Contradiction.

Where do I go wrong?

Solution 1:

You're imagining that the open intervals of $G^c$ are ordered discretely, like the integers, so you have alternating open intervals in $G^c$ and closed intervals in $G$. But actually, the open intervals of $G^c$ are densely ordered, and order-isomorphic to the rationals. As a result, there is no "next (...) in line for each [...]" like you claim there is. There are uncountably many closed intervals (actually, all of them are just single points) in between these open intervals, much like how there are uncountably many irrational numbers in between the rational numbers. There is no "next rational number" after each irrational number that you can use to get a bijection between rationals and irrationals, and the same thing is happening here.