Prove: $A\subseteq \mathbb{R}$ is an upper bounded set that contains at least two items/numbers. If $x < \sup{A}$, then $\sup{\left(A\setminus\{x\}\right)}=\sup A$.
My attempt:
Since $A$ is upper bounded, and non-empty, $A$ has a supremum according to the upper bound property, which we'll denote with $\sup{A}=c$. Since $c$ is $\sup{A}$, for all $a\in A$, $a< c$.
Therefore $\sup{A\setminus\{x\}}=\sup{A}$.
My questions is: Is my though process correct? I feel like I may not be understanding the question correctly, and this seemed to be too easy for an exam question.
Firstly, you made a little mistake in
It should be
To convince you, simply take the case where $\sup{A}\in A$ like in the case where $A=[1,2]$. Here $\sup{A}=2\in A$ and $\forall a\in A,\,a\le 2$.
Secondly, your conclusion
seems quite unclear to me. You didn't explain what you did with $c'=\sup{A\setminus\{x\}}$.
Thirdly, when you want to prove the equality of the $\sup$ of two different sets (here $c$ for one set as you named it and $c'$ for the other one as I named it) , I recommend you to, in general, try to prove that $c\le c'$ and that $c'\le c$. $c'\le c$ should be easy since $\left(A\setminus\{x\}\right)\subset A$. To prove $c\le c'$, you must clearly show how you used the assumption $x<\sup{A}$.