Cfrgtkky

Let $I,J$ be closed bounded intervals, and let $epsilon>0$ such that $l(I)> epsilon l(J)$. Assume that $I cap J neq emptyset,$ show that if $epsilongeq 1/2$, then $J subseteq 5 * I. $ Where $5* I$ denotes the interval with the samecenter as $I$ and five times its length. Is the same true if 0 < epsilon < 1/2?

My attempt:
Let $I=[a,b], J=[a,d]$ with $a,b,c,d$ are real numbers.
For $x in J$, we have $[c,x] subseteq J,$ so $1/2(x-c )leq epsilon (x-c) leq epsilon (d-c)<b-a $, so $x <frac{b+epsilon c}{epsilon}-frac{a}{epsilon}.$ I could not get more steps after this, so I would appreciate any comments or help with that, Thanks.

It's simply not true. Consider $J=[0,2]$, $I=[1,3]$. As $ell(I)=ell(J)$ the condition $ell(I)>epsilon,ell(J)$ is fulfilled with $epsilon={3over4}>{1over2}$. But $J$ is not contained in $5*I=[5,15]$.

I suggest you check your source. Maybe your interpretation of $5*I$ is wrong.

Assume that $I cap J= emptyset$ with $J,I$ are closed and bounded intervals, and let $gamma>0$ such that $l(I)>gamma. l(J)$. Let $y in J$, We have that $5*I=[x-frac{5}{2}l(I),x+frac{5}{2}l(I)]$ where $x$ is the center of $I$. To show that $y in 5 * I$, we will show that $|y-x|leq frac{5}{2} l(I).$ Since there exists $z in I cap J$, then by triangle inequality $$|y-x|leq |y+z|+|z-x|leq l(J)+l(I)<frac{1}{gamma}l(I)+frac{l(I)}{2}leq frac{5}{2}l(I)$$ for any $gamma geq 1/2,$ and we are done.
Now, consider $I=[0,1/2], J=[1/2,5]$. Clearly, $l(I)=1/2> gamma l(J)=gamma (4.5)$ where $gamma=1/10.$ We have that $5*I=[-1,frac{6}{4}]$, and clearly $J nsubseteq 5 *I $ since 4$notin J.$