0

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.

u123435
  • 173
  • 1
  • 5

2 Answers2

0

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...

cws
  • 474
0

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

Bram28
  • 100,612
  • 6
  • 70
  • 118
  • 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? – u123435 Jan 25 '17 at 03:53
  • @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. – Bram28 Jan 25 '17 at 04:32