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

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

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

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.

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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...

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

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    I think its possible for the statement to be false for when $A$ is infinite. However, if $A$ is finite, then $m$ can be the highest value, which would make the statement true. How would I go on after this? – 2017-01-25
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    @u123435 Exactly, it depends on the nature of $A$ whether the first claim is true or false. ... Though it is not quite right to say that $A$ must be finite in order for the claim to be true ... $A=[0,1]$ is infinite, but does have a highest value. Oh, and by the way, if the first claim is that there is a highest value for *any* $A$ then that is clearly false. For the second one: *if* there is a highest value in $A$, then yes, it has to be unique, for if there were two highest values, we reach a contradiction, since one has to be smaller than the other, and is therefore not a highest value. – 2017-01-25