$s + \frac{1}{n}$ is an upper bound for $A$ and $s - \frac{1}{n}$ is not an upper bound for $A$. Show $s = \sup A.$

Question

Let $A \subset \mathbb R$ be non empty and bounded above, and let $s \in \mathbb R$ have the property that for all $n \in \mathbb N$, $s + \frac{1}{n}$ is an upper bound for $A$ and $s - \frac{1}{n}$ is not an upper bound for $A$. Show $s = \sup A.$

I am having difficulty proving this statement. It is intuitively clear to me that it holds true but have no idea where to begin proving this. Thanks in advance for any help.

Answer 1

Pure definitions.

The sup must exist as the reals have the least upper bound property.

$s - \frac 1n$ is not an upper bound for all $n$ so $\sup A > s + \frac 1n$ or all natural $n$.

All $s + \frac 1n$ is an upper bound so $\sup A \le s+ \frac 1n$ for all natural $n$.

So $s-\frac 1n < \sup A \le s+ \frac 1n$ for all natural $n$.

Now either $\sup A < s$ or $\sup A = s$ or $\sup A > s$.

If $\sup A < s$ then $s - \sup A > 0$ and there is an $n$ in $\mathbb N$ so that $\frac 1n < s - \sup A$ an $s < s - \frac 1n$ which is a contradiction.

If $\sup A > s$ then $\sup A - s > 0$ and there is an $n$ so that $\frac 1n < \sup A -s$ and $s > \sup a+\frac 1n$ which is a contradiction.

So $\sup A = s$.

Of course you may need to review why it is true that for all $x > 0$ theren is a natural $n$ so that $0 < \frac 1n < x$....

Answer 2

For all $a\in A$, $a\leq s+\frac1n$ for all $n$. Then $a\leq s$ (if it weren't, if you had $s<a$, you would be able to squeeze, for $n$ big enough, $s<s+\frac1n<a$, contradicting that $s+\frac1n$ is an upper bound for $A$).

So $s$ is an upper bound for $A$. Given any $s'<s$, you can choose $n$ big enough so that $s'<s-\frac1n<s$. As $s-\frac1n$ is not an upper bound for $A$, there exists $a\in A$ such that $s-\frac1n<a$. Thus $$ s'<s-\frac1n<a, $$ showing that $s'$ is not an upper bound for $A$. In consequence, $s$ is the least upper bound of $A$: $s=\sup A$.

Answer 3

Since $s+1/n$ is an upper bound for $A$ we have $\sup A\leq s+1/n$ and further $s-1/n$ is not an upper bound so that there is an $a\in A$ which exceeds it and thus $s-1/n<a\leq \sup A$. It follows that we have $$s-\frac{1}{n}<\sup A\leq s+\frac{1}{n}$$ for all $n\in \mathbb {N} $. It is now almost obvious by Squeeze Theorem that $s=\sup A$.