Upper Bounds Of Integers Intersection

Question

Prove\Disprove:
$A$ is bounded from above $\iff$ $A\cap \mathbb{Z}$ is bounded from above.

Let $A=\{a\in \mathbb{Q} \setminus \mathbb{Z}: a<0\}$ is bounded from above, $A\cap \mathbb{Z}=\emptyset $ and $\emptyset$ is not bounded from above

Is it a valid contradiction?

Not every question involving sets is a set theory question. Elementary or otherwise. — Asaf Karagila, Jun 02 '15 at 18:00
E.g. $x<0$ is (vacuously) true for every $x\in\varnothing$. So $\varnothing$ is bounded from above. — drhab, Jun 02 '15 at 18:11

score 1 · Answer 1 · answered Jun 02 '15 at 18:09

I presume you are working in $\mathbb R$ or $\mathbb Q$.

If $s$ serves as upper bound of $A$ then it also serves as upper bound of any subset of $A$. So if $A$ is bounded from above then any subset of $A$, including $A\cap\mathbb Z$, is bounded from above as well.

If e.g. $A=\mathbb Q-\mathbb Z$ then $A$ is evidently not bounded from above while $A\cap\mathbb Z=\varnothing$ is bounded from above. So the converse of the statement is not true.

score 0 · Answer 2 · answered Jun 02 '15 at 18:07

0

The empty set is trivially bounded

answered Jun 02 '15 at 18:07

acradis

60

Upper Bounds Of Integers Intersection

2 Answers2