Supremum Exam Question

Question

I had a calculus exam and one of the questions (apparently an easy one ) had a lot of issues and our professor even emailed and said that most of the students didn't answer it correct with a formal proof.

I would like to ask for assistance of writing a formal proof for this question.

The question :

Let $A$ be a set containing at least 2 members. Prove that:

$\sup\{A\}>\sup\{A\setminus {x}\} \Rightarrow \sup\{A\}=x$

My try on this one:

Intuitively I understand why it is like that. Starting with negating the proposition 3

Answer 1

Yes, this can be proved by contradiction.

If $x \in A$ and $x < \sup A$, then there exists at least one $y \in A$ closer to $\sup A$ than $x$ is. (That's by the definition of $\sup A$.) Thus, we have $$ x < y \leq \sup A, $$ and, consequently, removing $x$ from $A$ does not affect the $\sup A$; i.e., $$ \sup(A) = \sup(A \setminus \{x\}). $$

Answer 2

I'd prove the contrapositive$\ldots$

Lemma: Suppose that $A$ is a nonempty subset of $\mathbb{R}$ which is bounded above, that $x\in A$, and that $x\neq\sup A$. In this case, $A\setminus\{x\}$ is nonempty and bounded above, and $\sup (A\setminus\{x\})=\sup A$.

Proof: $A\setminus\{x\}$ is nonempty because otherwise we'd have $A=\{x\}$ giving $\sup A=x$.

Since $A\setminus\{x\}\subset A$, we see that $A\setminus\{x\}$ is bounded above and $\sup(A\setminus\{x\})\leq\sup A$. (This is quick enough to prove.)

To show $\sup A\leq\sup(A\setminus\{x\})$ we need to show that $\sup(A\setminus\{x\})$ is an upper bound for $A$.

Take $a\in A$. Either $a\in A\setminus\{x\}$ or $a=x$. In the first case, $a\leq\sup(A\setminus\{x\})$ by definition of $\sup$. In the second we argue as follows. Since $a=x\neq\sup A$, we have $a<\sup A$ and we can find a $b\in A$ with $a<b$; we see $b\in A\setminus\{x\}$ so that $a<b\leq\sup(A\setminus\{x\})$, as required.