This is an example from Royden's Real Analysis 4th edition that I am having a lot of trouble understanding.
A countable set has outer measure zero. Let $C$ be a countable set enumerated as $C=\{c_k\}_{k=1}^{\infty}$. Let $\epsilon > 0$. For each natural number $k$, define $I_k=(c_k-\epsilon/2^{k+1}, c_k+\epsilon/2^{k+1})$. The countable collection of open intervals $\{I_k\}_{\infty}^{k=1}$ covers $C$. Therefore:
$$ 0 \leq m^*(C)\leq \sum^{\infty}_{k=1} \ell(I_k)=\sum^{\infty}_{k=1}\epsilon/2^k = \epsilon$$
This inequality holds for each $\epsilon > 0$. Hence $m^*(E) = 0$.
What I understand along with questions:
- A countable set is a set like the natural numbers.
- $C=\{c_k\}_{k=1}^{\infty}$ is just a way to describe said countable set using indexing for all of the elements in the set.
- $I_k=(c_k-\epsilon/2^{k+1}, c_k+\epsilon/2^{k+1})$ I kind of understand. This is defining an open cover over the countable set (I think).
- Why the $\epsilon/2^{k+1}$ instead of just $\epsilon$? It looks like the interval is continually shrinking as n goes to infinity.
- The equation in the middle is summing up all of the intervals. The outer measure is simply the sum of all of the open intervals in the set C. $m^*(C)$ is countably subadditive, hence the $\leq$ sign here: $m^*(C)\leq \sum^{\infty}_{k=1} \ell(I_k)$
- How is the last summation producing epsilon?
- How is this summing to zero?
Thank you for your help.
-Idle
Best Answer
6) $\displaystyle\sum_{k=1}^{\infty}\frac{\epsilon}{2^{k}}=\epsilon\sum_{k=1}^{\infty}\dfrac{1}{2^{k}}=\epsilon\cdot\dfrac{\dfrac{1}{2}}{1-\dfrac{1}{2}}=\epsilon$.
7) No, the sum is $\epsilon$, but now we have $0\leq m^{\ast}(C)<\epsilon$ for every $\epsilon>0$, if $m^{\ast}(C)>0$, then put $\epsilon$ to be $m^{\ast}(C)$, then we arrive to $m^{\ast}(C)<m^{\ast}(C)$, a contradiction.