In Royden and Fitzpatrick's Real Analysis, there's this part of proving that the Cantor Set is closed, uncountable and has measure zero that I don't get.
Here is how I interpret the first part:
$C_1 = [0,\frac{1}{3}] \bigcup [\frac{2}{3},1]$
$c_1$ is in $[0,\frac{1}{3}]$ xor $[\frac{2}{3},1]$.
Let $F_1$ be the interval which does not contain $c_1$.
That is, $F_1 = \begin{cases}
[0,\frac{1}{3}] & \text{ if } \ c_1 \notin [0,\frac{1}{3}] \\
[\frac{2}{3},1] & \text{ if } \ c_1 \notin [\frac{2}{3},1]
\end{cases}$
For the part where it says, "One of the two disjoint Cantor intervals in C2 whose union is F1 fails to contain the point c2", are there any intervals in C2 whose union is F1? There is no mention of this in the errata of the book so it's likely right.
From what I know, $C_2 = [0,\frac{1}{9}] \bigcup [\frac{2}{9},\frac{3}{9}] \bigcup [\frac{6}{9},\frac{7}{9}] \bigcup [\frac{8}{9},\frac{9}{9}]$, so it seems none of those intervals combine to $F_1$.
Could R&F have meant to say "One of the two disjoint Cantor intervals in C2 whose union is in F1"?
That is,
$F_2 = \begin{cases}
[0,\frac{1}{9}] \bigcup [\frac{2}{9},\frac{3}{9}] & \text{ if } c_2 \notin [0,\frac{1}{9}] \bigcup [\frac{2}{9},\frac{3}{9}]\\
[\frac{6}{9},\frac{7}{9}] \bigcup [\frac{8}{9},\frac{9}{9}] & \text{ if } c_2 \notin [\frac{6}{9},\frac{7}{9}] \bigcup [\frac{8}{9},\frac{9}{9}]
\end{cases}$
Best Answer
Yes. Thanks @DanielFischer Cantor Set is closed, uncountable and has measure zero