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 $n$ $$u-\frac{1}{n}<s_n$$

But also recall by definition of the supremum, $$s_n=s\leq\sup(S)$$ for all $s\in S$. Hence combining the inequalities $\implies$ $$u-\frac{1}{n}<s_n\leq\sup(S)\leq u+\frac{1}{n}\iff u-\frac{1}{n}\leq \sup(S)\leq u+\frac{1}{n}$$ 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 $$|\sup(S)-u|=0\iff u=\sup(S)$$


Is this okay?

  • 1
    $\begingroup$ 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". $\endgroup$ – JimmyK4542 Oct 9 at 18:04
  • $\begingroup$ The proof seems too long, but it is correct. $\endgroup$ – Matt Samuel Oct 9 at 18:09

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