Exercise 1.3.4

Assume that $A$ and $B$ are nonempty, bounded above, and satisfy $B\subseteq A$. Show
$\sup B \leq \sup A$.

Here's my shot:

Assuming the above conditions, also assume $\text{sup}(A)$ and $\text{sup}(B)$ exist, and denote them as $\beta_A$ and $\beta_B$. Thus $\beta_A \geq x$ for all $x \in A$. Since $B \subset A$, every element of $B$ is also less than or equal to $\beta_A$:

$$\beta_A \geq x \quad \forall x \in B.$$

So, $\beta_A$ is an upper bound of $B$. Thus it must be that $\beta_B \leq \beta_A$.

To see this, suppose that $\beta_B > \beta_A$. Since $\beta_A$ is also an upper bound for $B$, this violates the assertion that $\beta_B = \text{sup}(B)$, as it would fail to be the least upper bound.

Therefore,

$$\text{sup}(B) \leq \text{sup}(A).$$

This looks good - one comment. You don't need to "assume $\sup(A)$ and $\sup(B)$ exist". They do exist by the completeness axiom. So assert they exist and support your assertion with the completeness axiom.

Oh! I had thought about that, but wasn't sure whether $A$ and $B$ were subsets of real numbers or just of a general ordered field.