# Set of integers bounded above

Could you please check the validity of my solution to the following question?

Example: Show that if a set $A$ of integers is bounded above, then $A$ has a largest element.

My attempt is as follows: If $A$ is bounded above it has a supremum, say $M$. I will show that $M$ has to be contained by $A$.

Assume it is not so, namely $M \notin A$.

There has to be a number $m\in A$ such that $M-1<m<M$, otherwise $M-1$ would be an upper bound of $A$.

Additionally, there has to be another number $n\in A$ such that $m<n<M$, otherwise $m$ would be an upper bound of $A$.

Now, we can infer the following result,

$-n>-M$ and $M-1<m$ $\Rightarrow$ $0<n-m<1$.

Because $n$ and $m$ are integers, $n-m$ must be an integer. Therefore, the assumption in advance leads to a contradiction and $M$ must be contained by $A$.

• I think you can stop at $M-1<m<M$. How could you find an integer between $M-1$ and $M$? Commented Jun 11, 2015 at 20:30
• It has to be because otherwise $M-1$ will be an upper bound. Is not it right? Commented Jun 11, 2015 at 20:31
• Yes, and that's your contradiction right there Commented Jun 11, 2015 at 20:31
• I still do not understand. Commented Jun 11, 2015 at 20:35
• The original proof was good. No reason to change it, $M$ "could" be $88.88$. Commented Jun 11, 2015 at 20:36