2
$\begingroup$

I was reading Cantor set from Royden Real Anlysis and i have a doubt on this line marked in brackets , unable to identify the $F_{i}$ sets ,Any help!

here is the Proof attached

  • 0
    One of the two disjoint intervals whose union is $C_{1}$ can be $[0,\frac{1}{3}] or [\frac{2}{3},1]$ fails to contain $c_{1}$ , what can be $c_{1} \in \textbf{C}$? – 2017-02-20
  • 0
    oh i see i think $\textbf{C}$ always contains 0 and 1 – 2017-02-20
  • 0
    $c_1$ is any element of $C$, When we have disjoint sets, at most one of them contains $c_1$, or in other words at least one of the does **not** contain $c_1$. – 2017-02-20
  • 0
    But if i take $F_{1} = [0,1/3]$ and $c_{1} = 1$ then it works but fails when it says that "one of the two disjoint intervals in 2nd step whose union is $F_{1}$" but here it is not equal to $[0,1/3]$,so my choice is incorrect! – 2017-02-20
  • 0
    What fails? $c_2$ is some elements of $\bf{C}$, and $F_1$, in your case, if the union of $[0,1/9]$ and $[2/9,1/3]$. One of these sets, which we call $F_2$, does not contain $c_2$, because they are disjoint. The procedure described in the proof can still be done. – 2017-02-20
  • 0
    Yes,but how Union of $[0,1/9]$ and $[2/9,1/3]$ equal $[0,1/3] = F_{1}$? – 2017-02-20
  • 1
    You are right, that is a typo. After having $F_n$, $F_{n+1}$ should be defined as follows: "In the $n+1$-th step, one of the two disjoint intervals whose union **is contained in** $F_n$ does not contain $c_{n+1}$; Denote it by $F_{n+1}$." – 2017-02-20

0 Answers 0