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I want to prove the following statement using the theorem of Completeness of Numbers:

Suppose $S$ is a non-empty subset of real numbers. $S$ is an interval, if and only if for each pair of $x$ and $y$ in $S$ which $x < y$, if $z$ is a number that $x \leq z\leq y$, then $z$ is a member of $S$.

How can I achieve this?

Floris Claassens
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Jigsaw
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    Have you made an attempt? Did you try the 'only if part'? – Kavi Rama Murthy Oct 02 '19 at 08:04
  • Yes, in fact I’ve solved the first way like this: S is a interval of sth like [a,b], then both x and y are between a and b, so when z >= x and x >= a then z>=a and when z<=y and y<= b then z<=b and finally a <= z <= b and so z is included in the interval. My problems are about 1) the other way of the proof; 2) a better way to do the first way of the proof; and 3) how to apply the theorem of Completeness of numbers to it. – Jigsaw Oct 02 '19 at 08:10

2 Answers2

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Let $S$ be a subset of the real numbers such that for all $x,y\in S$ and $z\in\mathbb{R}$ we have that if $x<z<y$, then $z\in S$.

We want to show that $S$ is an interval, i.e., we need to find some $a,b\in\mathbb{R}$ with $a<b$ such that for all $z\in\mathbb{R}$ we have that if $a<z<b$ then $z\in S$ and if $z<a$ or $b<z$ then $z\not\in S$. (Note this way $S$ can be an open, closed or half-open interval.)

We have two clear candidates for $a$ and $b$, take $a=\inf S$ and $b=\sup S$, these numbers exists because $\mathbb{R}$ is complete. Note that we can have that $a=-\infty$ or $b=\infty$, unless $S$ is bounded. Now let $z\in\mathbb{R}$ such that $a<z<b$. As $a$ is the infimum of $S$ we can find some $c\in S\cap (a,z)$ and since $b$ is the supremum of $S$ we can find some $d\in S\cap (z,b)$. Then $c<z<d$ and by the properties of $S$ we find $z\in S$.

If $z<a$, then, as $a$ is a lower bound of $S$, clearly $z\not\in S$ and similarly if $b<z$, then, as $b$ is an upper bound of $S$ clearly $z\not\in S$. This was all we needed to show that $S$ is an interval.

Floris Claassens
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If $S$ is an interval (of any of the types), it's clear that it is order-convex (which is what the property defined by the statement about all $x<y$ etc. is called). You can check this by the definitions of the interval set and the transitivity of the order. No completeness is yet needed. (E.g. if $S=(a,b)$ and $x < y$ are in $S$ and $x \le z \le y$ then $a < x \le z \le y < b$ (the first and last come from $x,y \in S$) and then by transitivity $a < z < b$ and so $z \in S$, to see one case, the others are similar. Ctrl+C, Ctrl+V are your friend).

Now if $S$ is order convex, there are cases:

  • $S$ is bounded above, in that case let $b=\sup(S)$ which exists by completeness.
  • $S$ is not bounded above, in that case let $b=+\infty$.
  • $S$ is bounded below, in that case $a=\inf(S)$, or
  • $S$ is not bounded below, in that case $a=-\infty$.

If $a,b$ are finite and $a \in S, b \in S$, show that $S=[a,b]$. ($x \in [a,b]$ then $a \le x \le b$ by definition, but $a,b \in S$ and order-convexity of $S$ then implies $z \in S$. And if $x \in S$, $x \le b=\sup(S)$ is trivial, as is $a = \inf(S) \le x$ as well, so $x \in [a,b]$ and both inclusions have been shown.)

If $a,b$ are finite and $a \in S, b \notin S$ show that $S=[a,b)$ (also an interval type). The proof is similar.

You can formulate and prove the other (finite) cases as well.

If $b=+\infty$ and $a$ is finite and $a \notin S$ then we need to show $S=(a,+\infty)$ or $S=(a,\rightarrow)$ (another common notation) or just $S=\{x\in \Bbb R: x > a\}$. All such arguments are very similar. Try them.

If both $a,b$ are infinite, of course $S=\Bbb R$ (or $S=\emptyset$ if we're very pedantic ($\inf(\emptyset)=+\infty$ and $\sup(\emptyset)=-\infty$).

Henno Brandsma
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