1
$\begingroup$

Form a sequence $(En)_{n\ge0}$ of subsets of $R$ as follows: $E_{0} = [0, 1]$; $E_{1}$ is obtained from $E_{0}$ by removing the open middle interval of length $1/5$, i.e. $E_{1} = [0,2/5]∪[3/5, 1]$; $E_2$ is obtained by removing the open middle intervals of length $1/25$ from each of the $2$ intervals forming $E_{1}$, i.e. $E_{2} = [0,9/50 ] ∪ [11/50 ,2/5] ∪ [3/5,39/50 ] ∪ [41/50 , 1]$; In general, $E_{n}$ is obtained by removing the open middle intervals of length $1/5^{n}$ from each of the $ 2^{n−1}$ intervals forming $E_{n−1}$.

Let $E =\bigcap_{n=1}^{\infty} E_n$.

I'm trying to find the relation between $n$ and the length the of each $2^n$ interval ,

The only relation I could find is $L_{n}=(5^{n}-\sum_{k=1}^{n}5^{n-k}2^{k-1})/10^{n}$

Can any one help me with better relation , this one is not helping me to prove $E$ has zero measure .

$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

They've pretty much already told you the main point: you start with measure $1$ and at the $n$th step you remove $2^{n-1}$ disjoint intervals each of which has measure $5^{-n}$. So the amount of measure that you remove is $\sum_{n=1}^\infty 2^{n-1} 5^{-n}$. This sum is easy to calculate using the geometric series.

$\endgroup$
4
  • $\begingroup$ Thanks.Could you please give me the sum of this series $\endgroup$
    – majduleen zeyadeh
    Commented Mar 10, 2018 at 0:03
  • $\begingroup$ @majduleenzeyadeh Do you know how to find the sum of a geometric series? If not, look here. $\endgroup$
    – Noah Schweber
    Commented Mar 10, 2018 at 0:06
  • $\begingroup$ @NoahSchweber When you are finished summing his series, could you solve some quadratic equations for me ? $\endgroup$
    – Rene Schipperus
    Commented Mar 10, 2018 at 0:56
  • 1
    $\begingroup$ @ReneSchipperus I don't think that's called for - I directed the OP to a place they could learn a relevant technique. (Or did you not check where the link goes to?) $\endgroup$
    – Noah Schweber
    Commented Mar 10, 2018 at 1:22
0
$\begingroup$

You are making the problem harder than it really is.

Note that by removing the middle fifth each time you are left with $4/5$ of what you had before.

Thus $$L_0=1, L_1=4/5, L_2=(4/5)^2, L_3=(4/5)^3,..., L_n=(4/5)^n$$

$\endgroup$
6
  • $\begingroup$ Do you mean we can calculate the measure using the limit of Ln?? $\endgroup$
    – majduleen zeyadeh
    Commented Mar 10, 2018 at 0:23
  • $\begingroup$ @majduleenzeyadeh By continuity of measure, yes. $\endgroup$
    – Ian
    Commented Mar 10, 2018 at 0:26
  • $\begingroup$ but im not removing the middle fifth , im removing the interval of length $1/5^{n}$ $\endgroup$
    – majduleen zeyadeh
    Commented Mar 10, 2018 at 0:34
  • $\begingroup$ @majduleenzeyadeh No, at each stage you remove the middle fifth of all the intervals that are remaining. This will have a fifth of the total length of everything that is remaining. $\endgroup$
    – Ian
    Commented Mar 10, 2018 at 1:11
  • $\begingroup$ Yes I get it , just finished it , thanks all for your help . $\endgroup$
    – majduleen zeyadeh
    Commented Mar 10, 2018 at 2:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .