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 $\mathop{\mathrm{sup}} X \le \mathop{\mathrm{sup}} Y$
Best Answer
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..