TITLE: Length of intervals
QUESTION [0 upvotes]: 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 $\epsilon\geq 1/2$, 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.

REPLY [0 votes]: 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 $5I=[-1,\frac{6}{4}]= [-1, \frac 32]$, and clearly $J \nsubseteq 5I $ since $4\notin J.$
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