0
$\begingroup$

Working through foundations of mathematical anlysis by johnsonbaugh per suggestion and wondering if the following proof works? (no solutions to book)

Problem:

Let X and Y be nonempty subsets of real numbers such that $X \subset Y$ and Y is bounded above. Prove that $\mathop{\mathrm{sup}} X \le \mathop{\mathrm{sup}} Y$.

Proof:

Given that Y is bounded above, by the least upper bound axiom there exists a least upper bound of Y. Given that $X \subset Y$ we have that $(\forall x \in X)(x \in Y)$. Then, since Y is bounded above, X must be and we have the existence of a least upper bound for X. Now, by definition of a supremum for every upper bound, $a \in Y$, $a\ \le \mathop{\mathrm{sup}} Y$. Hence, for every upper bound $b \in X$, we must have that $b \le \mathop{\mathrm{sup}} Y$ and it follows that $\mathop{\mathrm{sup}} X \le \mathop{\mathrm{sup}} Y$

$\endgroup$
0

1 Answer 1

2
$\begingroup$

The last bit is confused. You know that $\sup(Y)$ exists and is an upper bound for $X$ as well, as $X \subseteq Y$. So $\sup(X)$ exists. But $\sup(X)$ is the smallest of all upperbounds of $X$, and $\sup(Y)$ is one of the possible upper bounds. So..

$\endgroup$
6
  • $\begingroup$ Not sure I follow $\mathop{\mathrm{sup}} Y$ is one of the possible upper bounds.. So, since $\mathop{\mathrm{sup}} X$ is the smallest of all upper bounds of X and $\mathop{\mathrm{sup}} Y$ is a possible upper bound it must be less than $\mathop{\mathrm{sup}} X$ by definition of $\mathop{\mathrm{sup}} X$ if im following? I thought that's what my last line stated? $\endgroup$
    – Pythonista
    Feb 6, 2016 at 14:25
  • $\begingroup$ Yes, but that was not what I read $\endgroup$ Feb 6, 2016 at 14:40
  • 1
    $\begingroup$ @Slayer $\sup Y$ is an upper bound for $Y$. Then $X\subseteq Y$ implies that it is an upper bound of $X$ as well. That implies directly that $\sup Y$ will not be smaller than the least upper bound of $X$. So $\sup X\leq \sup Y$. $\endgroup$
    – drhab
    Feb 6, 2016 at 14:40
  • $\begingroup$ You said "Now, by definition of a supremum for every upper bound, $a \in Y$, $a \le \sup Y$", which makes no sense. $\endgroup$ Feb 6, 2016 at 14:42
  • $\begingroup$ Okay now I see my error I didn't see it at first. That makes more sense. Thanks! $\endgroup$
    – Pythonista
    Feb 6, 2016 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.