2

Let $A \subseteq B$, where $A$ and $B$ are non-empty sets and $B$ is bounded above.

Show that $\sup(A)$ exists and $\sup(A)\le \sup(B)$ (Problem given as written)

For clarity, $\sup(x)$ is the supremum of x

I'm just asking if I did this correctly, I feel like I'm missing something here

Proof sup(A) exists: Let $Y = \max(B)$. Then, $Y \ge a, \forall a\in A$. Therefore, A is bounded above because $\exists b\in B$ such that $b \ge a,\forall a\in A$. By completeness axiom, if A is bounded above, then $\exists \sup(a)$

Proof $\sup(A)\le\sup(B)$: Since $\sup(B)$ is an upper bound of $B$ and $A\subseteq B$, $\sup(B)$ is an upper bound of $A$ such that $\sup(A) = \sup(B)$ or $\sup(A) < \sup(B)\implies \sup(A)\le \sup(B)$.

Angelo Rendina
  • 1,893
user3400223
  • 243

2 Answers2

1

Your first does not looks reasonable to me: for example $\max(B)$ may not exist. Instead try saying that

  • if $B$ is bounded above then there is some $y$ such that $\forall b \in B$ you have $b \le y$;
  • since $A \subseteq B$, $\forall a \in A$ you have $a \in B$ and so $a \le y$ and thus $A$ is bounded above;
  • so by completeness ...

Your second looks more reasonable to me. You could say that $\forall b \in B$ you have $b \le \sup(B)$ and so since $A \subseteq B$, $\forall a \in A$ you have $a \le \sup(B)$ and thus $\sup(A) \le \sup(B)$, but this is what I understood from what you said.

Henry
  • 157,058
1

The existence part is very close, but you can't let $Y$ be the maximum of $B$, because $B$ might not have a maximum! But it's bounded above, and if it is a subset of $\Bbb R$, then the completeness of $\Bbb R$ implies it has a least upper bound, i.e., $\sup{B}$ exists. If you let $Y = \sup{B}$ instead, then your argument for existence works perfectly. You show $Y$ is a bound for $A$, as you did, and then by completeness of $\Bbb R$, then $A$ is bounded from above and so has a least upper bound, i.e., $\sup{A}$ exists.

Your argument for the second part is also fine. Since $\sup{B}$ is an upper bound for $A$, and $\sup{A}$ is the least upper bound, it follows that $\sup{A} \leq \sup{B}$. Good job.

Just to add a comment: I wanted to give you an example of a subset of $\Bbb R$ that has a supremum but not a maximum. Take $A = (0, \pi) \cap \Bbb Q$. Clearly, this is bounded, so it has a least upper bound (which is actually $\pi$). But can you find a largest element in this set?

layman
  • 20,191