Let $E$ be a non-empty subset of $R$, let $n ≥ 1$ be an integer, and let $L<K$ be integers. Suppose that $K/n$ is an upper bound for $E$, but that $L/n$ is not an upper bound for $E$. Show that there exists an integer $L<m≤ K$ such that $m/n$ is an upper bound for $E$, but that $(m−1)/n$ is not an upper bound for $E$. (Hint: prove by contradiction, and use induction. It may also help to draw a picture of the situation.)
The picture: --------$x_{n-k-1}$-$L/n$-$x_{n-k}$...------$x_{n-1}$-$(m-1)/n$-$x_n$-$m/n$---$K/n$
Following the hint lets induct on $n$. The contradiction of the proposition will be that for any integer $L<m≤ K$ such that $m/n$ is not an upper bound for $E$, $(m−1)/n$ is an upper bound for $E$.
Then I think I should induct on $n$.
However, the contradiction of the proposition is itself faulty I think, because it dose not make any sense. Can you help me at least to start?
I have seen several discussions on the same question but none of them seems to me proceed the way the author suggested.
Suppose the result is false. Then, for any pair of successive integers $X$ and $X+1$, either both $X/n$ and ${(X+1)}/n$ are upper bounds or neither are upper bounds.
Then $L/n$ is not an upper bound and so $(L+1)/n$ is not an upper bound, $(L+1)/n$ is not an upper bound and so $(L+2)/n$ is not an upper bound, ...
By induction $N/n$ is not an upper bound for any positive integer $N$. This contradiction (for $N=K$) proves that the required result is true