I wanted to check if this proof works. Any comment would be appreciated!

Proof. Note that $\sup(A)$ is the smallest number such that $a \leq \sup(A)$ for every $a \in A$

Now it is true that $a \leq 1$ for all $a \in A$. Therefore $\sup(A) \leq 1$.

Now suppose $\sup(A) = b < 1$. Then $b \in A$

Let $\epsilon >0$ and b = $1 - \epsilon$. Then 1 - $\frac{\epsilon}{2}$ > b

So b can not be an upper bound of A since it can't be true that a $\le b$ for all a $\in A$

Hence we only have that $\sup(A) = 1$.

  • 1
    $\begingroup$ "...but then A has no maximum..." ?? $\endgroup$ – DonAntonio May 13 '16 at 21:10
  • $\begingroup$ @Joanpemo because 1 is not included in A, whatever that b is, wouldn't there be any number that is slightly greater than b? That was what I was thinking.. $\endgroup$ – Meer May 13 '16 at 21:19
  • $\begingroup$ That's something you have to prove, yet that wasn't my point but the word "then" there. $\endgroup$ – DonAntonio May 13 '16 at 21:21
  • $\begingroup$ @Joanpemo wording is wrong? $\endgroup$ – Meer May 13 '16 at 21:26
  • $\begingroup$ The logic is wrong! $\endgroup$ – The Chaz 2.0 May 13 '16 at 22:03

Take any $\;\epsilon>0\;$, (we can assume $\;\epsilon<1\;$ , otherwise the argument is trivial), and take $\;a:=1-\frac\epsilon2\;$ , then $\;a\in A\;$ and

$$1-\epsilon<a<1\,,\;\;\text{and since this is true for any}\;\;\epsilon >0$$

we get that $\;1=\sup A\;$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.