Show there exists an integer $LI'm trying to prove the following:
"Let $E$ be a non-empty subset of $\mathbb{R}$, let $n \geq 1$ 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 $E$. Without using the fact that if a non-empty subset of $\mathbb{R}$ has an upper bound then it must have exactly one least upper bound, show that there exists an integer $L<m\leq K$ 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.)"
My attempt:
We induct on $n$. First suppose that $n=1$. Then by hypothesis we have that $K$ is an upper bound for $E$ but $L$ is not. Suppose also that there is not an $L<m\leq K$ such that $m/n$ is an upper bound for $E$ but $(m-1)/n$ is not. But then if we take $m=K$ we have that $L<m\leq K$ and $m$ is an upper bound for $E$, a contradiction. So such an $m$ exists in the case $n=1$, as desired. Now suppose that the proposition is already proven for some $n$; we now prove it for $n+1$.
Here, the inductive step, is where I get stuck; I would appreciate any help about how to conclude this step.
Best regards,
lorenzo
NOTE: I'm aware that there are other questions on this site about this proposition, but none of them helped me finish the proof.
 A: There are a few caveats: We cannot assume that $L$ (or $K$) is $\ge0$, we only know that they are integers. Therefore I would suggest that you try to show that (under the assumption that the claim is false)
$$\tag {$1_k$} \frac{L+k}n\text{ is not an upper bound}$$
holds for all $k\in\mathbb N_0$.
We are already given that $(1_0)$ holds.
Assume $(1_k)$ holds. If $(1_{k+1})$ were false then $m=L+k+1$ would have the property asked for (or would it? Why does $L<m\le K$ follow?). As we assumed this false, we conclude $(1_{k+1})$ holds.
A: I think the question is looking too complicated only because of too much symbolism.
Let $r = K - L > 0$. Thus we the chain of integers $$L < L + 1 < L + 2 < \dots < L + r - 1 < L + r = K$$ and hence a chain or rationals $$\frac{L}{n} < \frac{L + 1}{n} < \dots < \frac{L + r - 1}{n} < \frac{L + r}{n} = \frac{K}{n}$$ and clearly there are $(r + 1)$ elements in this chain of rational numbers. Starting from the last member in the chain $K/n$ and going backwards check each element for being an upper bound for $E$. Clearly the last element in the chain is an upper bound for $E$ and first number of this chain is not an upper bound for $E$, it follows that there will be an element in the chain of type $(L + i)/n$ such that $(L + i)/n$ is an upper bound for $E$ and $(L + i - 1)/n$ is not an upper bound for $E$. The integer $m = L + i$ satisfies the conditions of the question.
Why do we need induction is not clear to me. Induction is not needed for reasoning with finite number of things, but rather for an infinite number of things.
