Solution verification: show that $s= \sup A$

Question

$A\subseteq\Bbb R, s\in\Bbb R$ s. t. $\forall n \ s+\frac1n$ is an upper bound of $A$ and $s-\frac1n$ isn't. Show that $s=\sup A$.

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My attempt:

Given $s - \frac1n$ isn't a lower bound of $A$. So $s -\frac1n< a,\ a\in A\implies A\ne\emptyset$. Also, from the hypothesis, $A$ is bounded above. So, $\exists t=\sup A$.

$t\geqslant a\ \forall a\in A$. From the hypothesis, for some $a$, we have $s-\frac1n\leqslant a$, so, we have, $-\frac1n\leqslant a\leqslant t$, which means $s-\frac1n\leqslant t$.

Since $t$ is the supremum of $A$ and $s +\frac1n$ is an upper bound, we have $t\leqslant s+\frac1n$.

Now we have, $s-\frac1n\leqslant t\leqslant s+\frac1n$, so, as $n\to\infty$, we get $s=t$.

Is my solution correct?

Thanks in advance.

Answer 1

Your proof is correct indeed, but I would try to avoid using limits to prove it. You can simply say that$$s-\frac1n\leqslant t\leqslant s+\frac1n\iff|t-s|\leqslant\frac1n.$$We can't have $|t-s|>0$, by the Archimedian property of the real numbers. So, since $|t-s|\geqslant0$, we have $|t-s|=0$. In other words, $t=s$.

Answer 2

Your proof is correct, but it can be done in a more straightforward way.

Let's show that $s$ is an upper bound of $A$. If not, there exists $a\in A$ such that $a>s$. Take $n$ such that $a-s>1/n$: then $a>s+1/n$, which contradicts $s+1/n$ being an upper bound of $A$.

Let's show that $s$ is the least upper bound of $A$. If not, there exists $t<s$ that's an upper bound. Take $n$ such that $s>t+1/n$: then $t<s-1/n$ and there exists $a\in A$ with $a>s-1/n$, contradicting the assumption that $t$ is an upper bound of $A$.

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Note that we don't even need to use the existence of the supremum: only the Archimedean property has been used.