Proof of the Cut Property using the Axiom of Completeness I'm reading Understanding Analysis by Stephen Abbott and wanted to ensure this proof I've just written is correct.
Here's the statement I'm proving (The Cut Property):
If $A,B$ are non-empty, disjoint sets so that $A \cup B = \mathbb{R}$, $A \cap B = \emptyset$, and $\forall a \in A, b \in B, a < b$ then $\exists c \in \mathbb{R}$ such that $\forall a \in A, b \in B, a \leq c \leq b$.
And here is my proof. I'm wondering if it is correct and if there is a simpler way I might have overlooked.
Proof. Since $A$ is non-empty and bounded above by any element of $B$ we know that $\sup{A}$ exists. We will take $c = \sup{A}$.
Since $A$ and $B$ are disjoint and contain all of the reals, $c$ must exist in one, but not both, of these sets. Will will split into cases from here.
Case $c \in A$.
From the definition of supremum we know that $a \leq c$ where $a \in A$. Additionally, since $c \in A$ we know from our assumption that $c < b$ where $b \in B$. Thus the original statement is proven in this case.
Case $c \in B$.
From our assumptions we know that $a < c$ where $a \in A$. To show that $c \leq b$ where $b \in B$ we will prove that $c = \inf{B}$. That is, we will show that $\forall \varepsilon > 0, \exists x \in B$ so that $c + \varepsilon > x$. Let $\varepsilon > 0$. Then take $x = c$ since  in this case $c \in B$. Then obviously $c + \varepsilon > c = x$. Thus $c = \inf{B}$.
In either case we have $\forall a \in A, b \in B, a \leq c \leq b$. Thus we have proven the cut property. $\square$
How does it look? Thanks!
 A: What you’ve done in the second case is not sufficient to show that $c\le b$ for every $b\in B$. To see this, suppose that $A=\{x\in\Bbb R:x\le 0\}$; clearly $c=0\notin B$, so this example falls in your second case. Now imagine that you lose track of what you’re doing and think that $c=1$. For each $\epsilon>0$ there is an $x\in B$ such that $1+\epsilon>x$, because we can always take $x=1$. According to your argument this would allow you to conclude that $1=\inf B$, but of course this is false: all that this shows is that
$$1=\inf\{x\in B:x>1\}\,,$$
not that $1=\inf B$. (This is a good thing, since $1$ isn’t $\inf B$.) Similarly, your argument shows only that $c=\inf\{x\in B:x>c\}$, not that $c=\inf B$: it does not rule out the possibility that some member of $B$ is less than $c$.
In your first case it is obvious that $c$ is a lower bound for $B$, so what still needs to be proved is that it is the greatest lower bound for $B$; you did that. In your second case it is obvious that if $c$ is a lower bound for $B$ at all, then it is the greatest lower bound, but you still have to prove that it is a lower bound for $B$. You proved that nothing larger than $c$ can be a lower bound for $B$, just as above I proved that nothing larger than $1$ can be a lower bound for $B$, but you did not prove that $c$ actually is a lower bound for $B$. (And your proof that nothing larger than $c$ can be a lower bound for $B$ is unnecessarily complicated: you could simply say that in this case $c\in B$, so of course $\inf B\le c$.)
You can actually handle both cases at once. Let $c=\sup A$, as you did, and suppose that there is a $b\in B$ such that $b<c$. By the definition of the supremum there is an $a\in A\cap(b,c)$. (If you prefer, let $\epsilon=c-b$, and choose $a\in A$ such that $c-\epsilon<a<c$; it’s just a slightly more complicated way of saying the same thing.) But then $b<a$, which is impossible. Thus, $c\le b$ for each $b\in B$, and the fact that $c=\sup A$ ensures that $a\le c$ for each $a\in A$.
