# Show $u=\sup(S)$ if $u+\frac{1}{n}$ is an upper bound and $u-\frac{1}{n}$ is a lower bound, for all $n$

Suppose $$S\subset R$$ is nonempty. Prove $$u=\sup(S)$$ if $$\forall n\in\mathbb{N}$$, $$u-\frac{1}{n}$$ is not an upper bound of $$S$$ and $$u+\frac{1}{n}$$ is an upper bound of $$S$$.

I was hoping someone could check my form of proof for validity:

Since $$u+\frac{1}{n}$$ is an upper bound of $$S$$, $$\sup(S)$$ exists in $$\mathbb{R}$$ and is such that $$\sup(S)\leq u+\frac{1}{n}$$

Similarly, since $$u-\frac{1}{n}$$ is not an upper bound $$\forall n$$, $$\exists s_n\in S$$ such that for every $$n$$ $$u-\frac{1}{n}

But also recall by definition of the supremum, $$s_n=s\leq\sup(S)$$ for all $$s\in S$$. Hence combining the inequalities $$\implies$$ $$u-\frac{1}{n} which further implies $$|\sup(S)-u|<\frac{1}{n}$$

By the Archimedean property, $$\forall \epsilon>0,$$ there exists a sufficiently large $$N$$ such that $$\frac{1}{N}<\epsilon$$. Hence $$\forall \epsilon>0$$ $$|\sup(S)-u|<\frac{1}{N}<\epsilon$$ thus $$|\sup(S)-u|=0\iff u=\sup(S)$$

QED.

Is this okay?

• FYI: Your title doesn't quite match your question. "$u - \tfrac{1}{n}$ is a lower bound" is not the same thing as "$u - \tfrac{1}{n}$ is not an upper bound". – JimmyK4542 Oct 9 at 18:04
• The proof seems too long, but it is correct. – Matt Samuel Oct 9 at 18:09