Question

Suppose that I have an infinite collection of open intervals \(O_n\) whose lengths are \(1/2^{n-1}\), where \(n\in\mathbb{N}\). Would then\[O=\bigcup_{n=1}^{\infty}O_{n}\]have a length of \(2\)? If so, does this imply that\[O^{c}=\bigcap_{n=1}^{\infty}O_{n}^{c}\]has infinite length?

Answer 1

i say that \(O\) would have length \(2\) because \(1+1/2+1/2^2+...=2\).

Answer 2

confused. you say the lengths are each
\[\frac{1}{2^{n-1}}\] but who says they are disjoint?

Answer 3

hmm very good point

Answer 4

suppose the sets are the intevals
\[(\frac{1}{2^{n-1}}, \frac{1}{2^{n-1}})\] then since they are nested the union would be the first (largest) one

Answer 5

i forgot to add that the open intervals are centered at unique rational numbers, but your point still stands.

Answer 6

i meant of course
\[(-\frac{1}{2^{n-1}}, \frac{1}{2^{n-1}})\]

Answer 7

i could argue that the greatest length the union of these intervals can attain is of 2, though.

Answer 8

actually i think it would be one

Answer 9

\[(-\frac{1}{2},\frac{1}{2})\] has length 1

Answer 10

that'd be the smallest (given they're all nested, like you described).

Answer 11

i think that would be the largest

Answer 12

if they're disjoint, then the total length would be 2.

Answer 13

wait... hmm...

Answer 14

yes if they were disjoint you could add them up as you did in a geometric series get 2

Answer 15

gotta run, good luck

Answer 16

kk thx