Suppose that S is a non-empty subset of $R$. Prove that is $M=supS$, then for every $\epsilon > 0$ there exists $s \in S$ such that $s>M - \epsilon$

Question

I'm taking a university real-analysis paper and I'm trying my hand at some proofs. Any feed back on my proof to the question above would be much appreciated!

Assume $SupS = M$ to be true. Therefore

1: $s \leq M$ for all $s \in S$

2: $s \leq K$ for all $s \in S$ $ \implies $ $M \leq K$

Take $s=M$ by (1), if $\epsilon > 0$ then $s>M-\epsilon$.

Now take $\delta > 0$ and $\delta < \epsilon$.

Then $s- \delta > M- \epsilon$.

Thanks!

Answer 1

You start with "take $s=M$", but a condition is that $s$ is an element of $S$. Be aware that we have no guarantee that $M\in S$.

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Solution:

Suppose that it is not true.

Then some $\epsilon>0$ exists such that $s\leq M-\epsilon$ for every $s\in S$.

That means that $M-\epsilon$ is an upper bound of the set $S$.

But $M-\epsilon<M$ so this contradicts that $M$ is the least upper bound.