# Two Versions of The Cut Property of the Real Numbers

I have been trying without succes to proove the cut property using the least upper bound property of the real numbers. It appears that there are two versions of the cut property:

If X and Y are nonempty subsets of R such that $$x < y$$ for all $$x \in X$$ and $$y \in Y$$, then there exists c ∈ R such that $$x \le c \leq y$$ for all and $$y \in Y$$

If X and Y are nonempty subsets of R such that $$x \leq y$$ for all $$x \in X$$ and $$y \in Y$$, then there exists c ∈ R such that $$x \leq c \leq y$$ for all and $$y \in Y$$

I only understand the proof for the second version of the cut property, without the strict inequality:

How are there these two versions? They can hardly be equivalent, but as the cut property is equvalent to the axiom of completeness this is a problem right? What am I missing here?

These two formulations are, in fact, equivalent.

Clearly the second implies the first. (If every element of $X$ is $<$ every element of $Y$, then clearly every element of $X$ is $\le$ every element of $Y$.)

To show the other direction, suppose the first version holds, and $X, Y$ are sets of real numbers such that every element of $X$ is $\le$ every element of $Y$. There are two cases.

• If there is no $a\in X\cap Y$, then every element of $X$ is in fact $<$ every element of $Y$; so the existence of the desired $c$ follows immediately from the first version of the cut principle.

• If there is some $a\in X\cap Y$, let $c=a$; I claim this $c$ has the desired properties. Let $x\in X$; then if $x>c$, since $c\in Y$ we would have a contradiction with the assumption "each element of $X$ is $\le$ each element of $Y$." So in fact $c\ge x$ for all $x\in X$. Similarly, we can show that $c\le y$ for all $y\in Y$.

So if every element of $X$ is $\le$ every element of $Y$, then - in either of the two cases - there is some $c$ which is $\ge$ every element of $X$ and $\le$ every element of $Y$.

So the two versions of the cut principle are in fact equivalent.

• So as a mathematical novice It only now occures to me, that in the answer I linked and both the definitions above $X \cap Y = \emptyset$ was not assumed. So we could just take the $x \leq y$ version of the statement and say if we assume $X \cap Y = \emptyset$ then $x \leq y$ is equivalent to $x<y$ right? So the second parts of your answer simply means if $X \cap Y \neq \emptyset$ is true then we need to write $x \leq y$ instead of $x<y$ am I correct? Jun 6, 2017 at 17:53
• @pindakaas Yes. If $X$ and $Y$ aren't disjoint, then they have a unique element in common and this element is the desired $c$; and otherwise $x\le y$ is equivalent to $x<y$ for $x, y\in X, Y$ respectively. Jun 6, 2017 at 17:56
• would the relation $x \leq c \leq y$ not hold for all $c \in X \cap Y$ so there could be more then one? Jun 6, 2017 at 18:00
• @pindakaas We can show that $X\cap Y$ has at most one element. For suppose $c, d\in X\cap Y$ are distinct. Then either $c<d$ or $d<c$; suppose without loss of generality that $c<d$. Then $c\in Y$, $d\in X$, but $d\not\le c$, contradicting our assumption on $X$ and $Y$. So $X\cap Y$ can have at most one element. Jun 6, 2017 at 18:03

Any member of $Y$ is an upper bound for $X,$ and $X,Y$ are not empty, so $lub (X)$ exists. Let $c=lub(X).$

Obviously $x\leq c$ for all $x\in X$ because $c$ is an upper bound for $X.$

For any $y\in Y$ we have $X\subset (-\infty,y]$, so $$c=lub(X)\leq lub ((-\infty,y])=y.$$ So $c\leq y$ for all $y\in Y.$ QED.

Remark: The lub property of $\mathbb R$ implies that $\mathbb R$ also has the glb property.

Proof: Let $\phi \ne S \subset \mathbb R$ such that $S$ has a lower bound. Let $L(S)$ be the set of all lower bounds for $S.$ Any $x\in S$ is an upper bound for $L(S),$ and $S\ne \phi$, so $lub (L(S))$ exists.

Now for any $x\in S$ we have $L(S))\subset (-\infty,x],$ which implies $$lub (L(S) \leq lub ((-\infty,x])=x.$$ So $lub (L(S))$ is a lower bound for $S$,which, by the definition of $L(S)$, means that $$lub(L(S))\in L(S).$$

For any $Y\subset \mathbb R,$ if $lub (Y)\in Y$ then $lub (Y)$ is the greatest member of $Y.$

Therefore $lub(L(S))$ is the greatest member of $L(S).$ That is, $lub (L(S))$ is the greatest lower bound for $S.$