Question about a conclusion of MIT's study materials 
conclusion: if $L \in \mathbb{N}$ and $L>b-a$ where $a\in \Bbb{R} , b\in \Bbb{R}$ and $b>a$, then exist an integer $k$ s.t $\frac{k}{L} \in [a,b]$.

But I can find an counter example (e.g a=1/3, b=1/2, L=1)
The Original link (Problem 5):https://ocw.mit.edu/courses/mathematics/18-100b-analysis-i-fall-2010/study-materials/MIT18_100BF10_pset9.pdf
The Original link (solution 5):https://ocw.mit.edu/courses/mathematics/18-100b-analysis-i-fall-2010/study-materials/MIT18_100BF10_pset9sol.pdf
Is it wrong? If it's wrong, how to fix this conclusion?
 A: Explanation to the comment above, as requested by the OP.

Given $a<b$, you are looking for an integer $L>0$ such that for some integer $k$, $\frac kL\in [a,b]$.

First an informal description of the problem: let's say you are given a "target" interval $I=[a,b]$, and you make successive "jumps" of length $\epsilon$, starting from $0$. The goal is to jump on $I$ at some point. You certainly can't jump "over" $[a,b]$ if $\epsilon\leq b-a$. So, you are sure you will attain the target if $\epsilon\leq b-a$. On the other hand, if $\epsilon> b-a$ and you are free to choose $a$ for a fixed value of $b-a$, then it is certainly possible to jump over $[a,b]$: just choose $a>0$, very close to $0$, so that you still have $b<\epsilon$. Therefore, $\epsilon\le b-a$ is a necessary and sufficient condition to find a $k$ such that $k\epsilon\in [a,b]$.
If the above does not look enough, then we can find the exact value of $k$.
The condition is equivalent to $aL\leq k\leq bL$, or $0\leq k-aL\leq (b-a)L$. Then, if $(b-a)L\geq1$, you can choose $k=\mathrm{ceil}(aL)$. Since $x\leq \mathrm{ceil}(x)\lt x+1$ for all $x$, we then have, as requested: $$0\leq \mathrm{ceil}(aL)-aL\lt1\le (b-a)L$$
If, however, $(b-a)L<1$, we can find $a$ and $b$ that will fail. Take for instance $a=\frac{1}{3L}$ and $b=\frac{1}{2L}$ (basically the OP example). Then


*

*for $k\leq0$, $\frac kL\le0\lt a$ hence $\frac kL\notin[a,b]$

*for $k\ge1$, $\frac kL\ge \frac1L\gt b$ hence $\frac kL\notin[a,b]$


So if $(b-a)L<1$, there are values of $a,b$ such that the requirement that $\frac kL\in[a,b]$ for some $k$ fails.

Now, I didn't investigate completely the proof in the original solution to your problem 5, but the obviously incorrect part is now fixed: if $L_0$ is the smallest integer such that $(b-a)L_0\geq1$, then the proof requires an $L$ such that $L\ge \frac nm$ and $L\ge L_0$, hence $(b-a)L\ge(b-a)L_0\ge1$ and this condition is still satisfied.
