Characterizing real intervals

Let $\mathbb{S} \subset \mathbb{R}$.

• Let $\mathbb{L}(\mathbb{S}) = \{ x \in \mathbb{R}: x\hspace{3pt} \textrm{is a lower bound for} \hspace{3pt}\mathbb{S}\}$.

Then $\mathbb{L}(\mathbb{S})$ is either empty or $\mathbb{R}$ or is the closed interval $(-\infty, \inf(\mathbb{S})]$.

• Let $\mathbb{U}(\mathbb{S}) = \{ y \in \mathbb{R}: y\hspace{3pt} \textrm{is an upper bound for} \hspace{3pt}\mathbb{S}\}$.

Then $\mathbb{U}(\mathbb{S})$ is either empty or is $\mathbb{R}$ or is the closed interval $[\sup(\mathbb{S}), \infty)$.

• Let $\mathbb{Z}(\mathbb{S}) = \{ z \in \mathbb{R}: z\hspace{3pt} \textrm{is neither an upper bound nor a lower bound for} \hspace{3pt}\mathbb{S}\}$.

Note that the set $\mathbb{Z}(\mathbb{S})$ is empty (which is counted as an open set) if and only if $\mathbb{S}$ has at most one element.
Specifically we have:

• $\mathbb{L}(\phi) = \mathbb{U}(\phi) = \mathbb{R}$, $\mathbb{Z}(\phi) = \phi$.
• If $\mathbb{S} = \{s\}$ for some $s \in \mathbb{R}$, then $\mathbb{L}(\mathbb{S}) = (-\infty, s]$, $\mathbb{U}(\mathbb{S}) = [s, \infty)$ and $\mathbb{Z}(\mathbb{S}) = \phi$.

When $\mathbb{S}$ has at least two elements, the sets $\mathbb{L}(\mathbb{S})$, $\mathbb{U}(\mathbb{S})$ and $\mathbb{Z}(\mathbb{S})$ are pairwise disjoint and have union $\mathbb{R}$.
Further $\mathbb{Z}(\mathbb{S})$, being the complement of the union of $\mathbb{L}(\mathbb{S})$ and $\mathbb{U}(\mathbb{S})$ is a non-empty open interval.
In particular, the set $\mathbb{Z}(\mathbb{S})$ is always open, for any $\mathbb{S} \subset \mathbb{R}$. Using the concept $\mathbb{Z}(\mathbb{S})$, we have the following theorem for a subset $\mathbb{S}$ of the reals.

• The following are equivalent:
  • $\mathbb{Z}(\mathbb{S}) \subset \mathbb{S}$
  • If $x \in \mathbb{S}$ and $y \in \mathbb{S}$ with $x \le y$, then any $z \in \mathbb{R}$ with $x \le z \le y$ lies in the set $\mathbb{S}$.

Proof:

• If $\mathbb{S}$ is the empty set, $\phi$, then both conditions hold, since then $\mathbb{Z}(\mathbb{S}) = \phi = \mathbb{S}$ and the second condition holds vacuously.

So henceforth, we may assume that $\mathbb{S}$ is non-empty.

• First suppose that a non-empty set $\mathbb{S}$ of reals has the property that whenever $x \in \mathbb{S}$ and $y \in \mathbb{S}$, then any real $z$ with $x \le z \le y$ lies in $\mathbb{S}$.
  We prove that $\mathbb{Z}(\mathbb{S}) \subset \mathbb{S}$.
  
  So let $z \in \mathbb{Z}(\mathbb{S})$.
  Then, by definition, $z$ is neither an upper bound nor a lower bound for the set $\mathbb{S}$.
  • Since $z$ is not an upper bound for $\mathbb{S}$, an element $y$ of $\mathbb{S}$ must exist, with $z > y$.
  • Since $z$ is not a lower bound for $\mathbb{S}$, an element $x$ of $\mathbb{S}$ exists, with $x < z$.
  Then we have $x < y < z$ and $x \in \mathbb{S}$ and $y \in \mathbb{S}$, so $z \in \mathbb{S}$.
  So $\mathbb{Z}(\mathbb{S}) \subset \mathbb{S}$ and we are done.
• Conversely, let $\mathbb{S}$ be a non-empty set with $\mathbb{Z}(\mathbb{S}) \subset \mathbb{S}$.
  Let $x$ and $y$ in $\mathbb{S}$ be given, with $x \le y$.
  Let $z \in \mathbb{R}$ obey $x \le z \le y$.
  We must prove that $z \in \mathbb{S}$.
  
  If $z = x$ or $z = y$, there is nothing to do, since then $z \in \mathbb{S}$.
  So we may assume that $x < z < y$.
  • Then $z$ is not an upper bound for $\mathbb{S}$, since $z < y$ and $y \in \mathbb{S}$.
  • Also $z$ is not a lower bound for $\mathbb{S}$, since $x < z$ and $x \in \mathbb{S}$.
  Since $z$ is neither an upper bound nor a lower bound for the set $\mathbb{S}$, we have $z \in \mathbb{Z}(\mathbb{S})$ by definition of $\mathbb{Z}(\mathbb{S})$, so $z \in \mathbb{Z}(\mathbb{S})\subset \mathbb{S}$, so $z \in \mathbb{S}$ and we are done.

Now we use the theorem just proved to define an interval.

• A subset $\mathbb{S}$ of the reals is said to be an interval if and only if either of the following equivalent conditions holds:
  • $\mathbb{Z}(\mathbb{S}) \subset \mathbb{S}$.
  • If $x \in \mathbb{S}$ and $y \in \mathbb{S}$ with $x \le y$, then any $z \in \mathbb{R}$ with $x \le z \le y$ lies in the set $\mathbb{S}$.

We want to determine all possible intervals.

We have the following classification theorem.

• Let $\mathbb{S}$ be an interval (as just defined).

Then $\mathbb{S}$ is one of the following ten types of sets:

• $\phi$.
  This is said to be open and closed and $\mathbb{Z}(\phi) = \phi$.
• $\mathbb{R} = (-\infty, \infty)$.
  This is said to be both open and closed and $\mathbb{Z}(\mathbb{R}) = \mathbb{R}$.

The open non-closed intervals:

• $(-\infty, b) = \{ x \in \mathbb{R}: x < b\}$ for some $b \in \mathbb{R}$.
  This is said to be open and not closed and $\mathbb{Z}((-\infty, b)) = (-\infty, b)$.
• $(a, \infty) = \{ x \in \mathbb{R}: x > a\}$ for some $a \in \mathbb{R}$.
  This is said to be open and not closed and $\mathbb{Z}((a, \infty)) = (a, \infty)$.
• $(a, b) = \{ x \in \mathbb{R}: a < x < b\}$, for some $a$ and $b$ in $\mathbb{R}$, with $a < b$.
  This is said to be open and not closed and $\mathbb{Z}((a, b)) = (a, b)$.

The closed non-open intervals:

• $(-\infty, b] = \{ x \in \mathbb{R}: x \le b\}$ for some $b \in \mathbb{R}$.
  This is said to be closed and not open and $\mathbb{Z}((-\infty, b]) = (-\infty, b)$.
• $[a, \infty) = \{ x \in \mathbb{R}: x \ge a\}$ for some $a \in \mathbb{R}$.
  This is said to be closed and not open and $\mathbb{Z}([a, \infty)) = (a, \infty)$.
• $[a, b] = \{ x \in \mathbb{R}: a\le x \le b\}$, for some $a$ and $b$ in $\mathbb{R}$, with $a \le b$.
  This is said to be closed and not open.
  • In the case that $a = b$, the interval is $[a, a]$ and consists of a single point $a \in \mathbb{R}$ and then $\mathbb{Z}([a, a]) = \phi$.
  • Otherwise we have $a < b$ and then $\mathbb{Z}([a, b]) = (a, b)$.

The intervals that are neither closed nor open:

• $(a, b]= \{ x \in \mathbb{R}: a < x \le b\}$, for some $a$ and $b$ in $\mathbb{R}$, with $a < b$.
  This is said to be neither open nor closed and $\mathbb{Z}((a, b]) = (a, b)$.
• $[a, b) = \{ x \in \mathbb{R}: a\le x < b\}$, for some $a$ and $b$ in $\mathbb{R}$, with $a < b$.
  This is said to be neither open nor closed and $\mathbb{Z}([a, b)) = (a, b)$.

We notice that if $\mathbb{S}$ is any one of these ten kinds of sets, we always have $\mathbb{Z}(\mathbb{S}) \subset \mathbb{S}$ and in all these ten cases, the set $\mathbb{Z}(\mathbb{S})$ is the interior of $\mathbb{S}$: this is, by definition, the largest open subset of $\mathbb{S}$.

Proof of the classification theorem.
We need to prove:

• If $\mathbb{S}$ is a subset of the reals, such that $\mathbb{Z}(\mathbb{S}) \subset \mathbb{S}$, then $\mathbb{S}$ is one of the ten kinds of sets discussed above.

Proof:

• First, if $\mathbb{S} = \phi$, then $\phi = \mathbb{Z}(\mathbb{S}) \subset \mathbb{S} = \phi$ is true and the set $\phi$ is in our list above, so we are done.

So henceforth we may assume that $\mathbb{S}$ is non-empty.

• If $\mathbb{S}$ has no upper or lower bound, then any real $z$ is neither an upper bound nor a lower bound for the set $\mathbb{S}$, so $\mathbb{R} = \mathbb{Z}(\mathbb{S}) \subset \mathbb{S}\subset \mathbb{R}$.
  So we have $\mathbb{S} = \mathbb{R}$.
• If $\mathbb{S}$ is bounded below, but not bounded above, then $a = \inf(\mathbb{S})$ exists.
  Also $\mathbb{S} \subset [a, \infty)$, by definition of $a$.
  Then any real $z > a$ is neither an upper bound nor a lower bound for the set $\mathbb{S}$, so $(a, \infty) \subset \mathbb{Z}(\mathbb{S}) \subset \mathbb{S}\subset [a, \infty) = (a, \infty) \cup \{a\}$.
  So we have $\mathbb{S} = (a, \infty)$ or $[a, \infty)$.
• If $\mathbb{S}$ is bounded above, but not below, then $b = \sup(\mathbb{S})$ exists.
  Also $\mathbb{S} \subset (-\infty, b]$, by definition of $b$.
  Then any real $z < b$ is neither an upper bound nor a lower bound for the set $\mathbb{S}$, so $(-\infty, b) \subset \mathbb{Z}(\mathbb{S}) \subset \mathbb{S}\subset (-\infty, b] = (-\infty, b) \cup\{b\}$.
  So $\mathbb{S} = (- \infty, b)$ or $(- \infty,b]$.
• If $\mathbb{S}$ is bounded above and below, then $b = \sup(\mathbb{S})$ exists, as does $a = \inf(\mathbb{S})$ and we have $a \le b$.
  Then $\mathbb{S}\subset [a, b]$, by the definitions of $a$ and $b$.
  • If now $a = b$, we have $\mathbb{S} = \{a\}$, a closed interval.
    So henceforth we may assume that $a \ne b$, so $a < b$.
  • Then any real $z$, such that $a < z < b$ is neither an upper nor a lower bound for the set $\mathbb{S}$, so we get that:
    $(a, b) \subset \mathbb{Z}(\mathbb{S}) \subset \mathbb{S}\subset [a, b] = (a, b) \cup \{a, b\}$.
    So we conclude that the set $\mathbb{S} = (a, b)$ or $(a,b]$ or $[a, b)$ or $[a, b]$.

In all cases, we have shown that $\mathbb{S}$ is one of the kinds of sets in our list and the theorem is proved. We see also that the cases that turn up cover all ten possible kinds of non-empty intervals, as expected. Finally note the corollary:

• $\mathbb{S} \subset \mathbb{R}$ is an open interval if and only if $\mathbb{Z}(\mathbb{S}) = \mathbb{S}$.

If we do this, then we see that the complement of any interval is either an interval or a union of two intervals and the complement of an open interval is either a closed interval or the union of two closed intervals, whereas the complement of a closed interval is either an open interval or the union of two open intervals.

In general we define:

• An open subset of the reals is a subset that is a union of open intervals.
• A closed subset of the reals is a subset that is the complement of an open set.

The open and closed subsets form a topology for the reals:

• $\phi$ and $\mathbb{R}$ are open.
• The union of any collection of open sets is open.
• The intersection of any finite collection of open sets is open.

Phrased in terms of closed sets the topology obeys:

• $\phi$ and $\mathbb{R}$ are closed.
• The intersection of any collection of closed sets is closed.
• The union of any finite collection of closed sets is closed.