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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.

user13
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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

  • Why is that true that either both X/n and (X+1)/n are upper bounds or neither are upper bounds? Also by default it is given that L/n is not an upper bound. How dose that follow from it that the (L+1)/n, (L+2)/n... are also not upper bounds? – user13 Aug 26 '19 at 11:26
  • We are assuming the result is false and so we cannot have X/n not an upper bound but (X+1)/n an upper bound. –  Aug 26 '19 at 11:35