$\exists p \in A$, $\forall q \in A $ , $q\leq p$

Question

If $A \subseteq \mathbb{R} $ $$ \exists p \in A, \forall q \in A , q \leq p $$

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Can I just use a specific value for $p$ and arbritary value for $q$ to disprove this?

$p = 3$ and $q = p + 1$, hence $q > p$

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Also, how would should one go about this one:

If .. $\exists p \in A, \forall q \in A , q \leq p $ .. then $p$ is unique.

Answer 1

Hints:

For the first one, try a proof by contradiction. Assume it is true for some $p$, and show that there is a $q$ which gives a contradiction. It is not enough to give an example because maybe there is a number $p\in A$ that you did not consider.

Edit: as mentioned in another other answer, an appropriate choice of $A$ should be made. Since $A\subseteq \mathbb{R}$ (and not just $A\subset \mathbb{R}$), you can make $A=\mathbb{R}$.

For the second, also try a proof by contradiction. In this case, assume $p$ is not unique and see what follows...

Answer 2

For the first one, you need to define $A$ if you want to provide a counterexample ... HINT what if $A=\mathbb{R}$?