3
$\begingroup$

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!

$\endgroup$

1 Answer 1

3
$\begingroup$

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 $a\in A$.

$\endgroup$
4
  • $\begingroup$ Is there a typo in the first paragraph or am I miss reading? $0 \notin B$ since $0 \in A$? $\endgroup$
    – Derek C.
    Commented May 1, 2021 at 0:53
  • $\begingroup$ @DerekC.:Oops! Definitely a typo. $\endgroup$ Commented May 1, 2021 at 1:47
  • $\begingroup$ Ok one more things to discuss as I'd like to understand this. Why the $d=1$? the supremum of $A$ is 0 and thus we would have the first case. I understand that the usage of epsilon I've done is wrong. $\endgroup$
    – Derek C.
    Commented May 2, 2021 at 0:54
  • 1
    $\begingroup$ @DerekC.: I could have used any positive real number instead of $1$; $1$ was just easy to write. I was demonstrating that your argument simply doesn’t prove what you claimed, because it applies just as well to $1$ as to the true value, $0$, of $\inf B$. I’ve expanded that explanation; see if it helps. $\endgroup$ Commented May 2, 2021 at 1:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .