I am trying to write down a proof of the following fact:
Let $I$ and $J$ be non-empty intervals of $\mathbb{R}$ such that $\inf I \leqslant \inf J$. The union $I \cup J$ is an interval if and only if one of the following conditions holds:
- $I \cap J \neq \emptyset$;
- $\sup I = \inf J$, and $\sup I = \inf J \in I \cup J$.
Sufficiency
First, we show that, if $I \cap J \neq \emptyset$, then $I \cup J$ is an interval. Let $x, y \in I \cup J$ and $z \in \mathbb{R}$ such that $x \leqslant z \leqslant y$, and let $t$ be an arbitrary element of $I \cap J$. We consider three cases, in order to show that $z \in I \cup J$:
If $z = t$, then $z \in I \cap J \subseteq I \cup J$.
If $z > t$, then $t < z \leqslant y$. Therefore, if $y \in I$, then $z \in I$, because $t, y \in I$ and $I$ is an interval; if $y \in J$, then $z \in J$, because $t, y \in J$ and $J$ is an interval. Hence, $z \in I \cup J$.
If $z < t$, then $x \leqslant z < t$. Therefore, if $x \in I$, then $z \in I$, because $t, x \in I$ and $I$ is an interval; if $x \in J$, then $z \in J$, because $t, x \in J$ and $J$ is an interval. Hence, $z \in I \cup J$.
Next, we show that, if the second condition holds, then $I \cup J$ is an interval. Call $a$ the least upper bound of $I$, which is equal to the greatest lower bound of $J$ (by hypothesis). Let $x, y \in I \cup J$ and $z \in \mathbb{R}$ such that $x \leqslant z \leqslant y$. We consider three cases, in order to show that $z \in I \cup J$:
- If $z = a$, then $z \in I \cup J$ by hypothesis.
- If $z < a$, then $x \in I$, because $x \leqslant z < a$, while $n \geqslant a\ \forall \, n \in J$. By definition of least upper bound, there exists $u \in I$ such that $u > z$. Therefore, we have $x \leqslant z < u$, and $z \in I$ because $I$ is an interval. Hence, $z \in I \cup J$.
- If $z > a$, then $y \in J$, because $a < z \leqslant y$, while $m \leqslant a\ \forall \, m \in I$. By definition of greatest lower bound, there exists $v \in J$ such that $v < z$. Therefore, we have $v < z \leqslant y$, and $z \in J$ because $J$ is an interval. Hence, $z \in I \cup J$.
Necessity
Suppose $I \cap J = \emptyset$. We show by contradiction that the second condition holds.
Suppose $\inf J < \sup I$. If $\inf J = \sup J$, then $J$ consists of only one point. It follows from the definition of interval that $I \supseteq\ ]\inf I, \sup I[$. So, it would be $J \subseteq I$, which is absurd (because $I \cap J = \emptyset$). Therefore, $\inf J < \sup J$. Call $m$ the minimum of $\sup I$ and $\sup J$. We have \begin{equation*} \inf I \leqslant \inf J < \frac{1}{2} (\inf J + m) < m. \end{equation*} Since $I \supseteq\ ]\inf I, \sup I[$ and $J \supseteq\ ]\inf J, \sup J[$ (this follows from the definition of interval), we have $\frac{1}{2} (\inf J + m) \in I \cap J$, which is absurd.
Suppose $\sup I < \inf J$. Let $x$ be an arbitrary element of $I$, $y$ an arbitrary element of $J$ and $z := \frac{1}{2}(\sup I + \inf J)$. Then we have \begin{equation*} x \leqslant \sup I < z < \inf J \leqslant y. \end{equation*} $I \cup J$ is an interval, so $z \in I \cup J$, which is absurd because $z \notin I$ (since $z > \sup I$) and $z \notin J$ (since $z < \inf J$).
Therefore, $\sup I = \inf J$. Call it $a$. We observe that $a \in \mathbb{R}$, otherwise it would be $a = +\infty$ and $J \subseteq I$, which is absurd.
Finally, we show that $a \in I \cup J$. Let $x$ be an arbitrary element of $I$ and $y$ an arbitrary element of $J$. Then we have
\begin{equation*}
x \leqslant a \leqslant y.
\end{equation*}
Therefore, $a \in I \cup J$, because $I \cup J$ is an interval.
Is this proof correct? If it is, can it be improved? Are there other proofs of this fact?
Edited: The OP commented on a gap in an earlier draft of this proof in the situation that $I$ or $J$ degenerates to a single point.
However, with only the hypothesis "$\inf(I) \le \inf(J)$", if one allows degenerate intervals then the main statement in the question is false. A counterexample is $J = \{p\}$ and $p = \inf(J)=\sup(J)=\inf(I) \not\in I$. In that case $I \cup J$ is an interval, $I \cap J = \emptyset$, and $\sup(I) > \inf(J)$.
One way to repair this would be to strengthen the hypothesis
by replacing it with
I have rewritten to proof to work under this stronger hypothesis.
For sufficiency, there is a somewhat slicker proof by contradiction.
Consider first the case that $\inf(I) = \inf(J) = \{p\}$. Set $q = \sup(I) \le r = \sup(J)$. It follows that $I \cup J$ is an interval with endpoints $p,r$ (possibly missing one or both endpoints in the case that $p<r$; and equaling $\{p\}$ in the case that $p=r$). If $p<q$ then $I \cap J$ contains $(p,q)$ and so $I \cap J \ne \emptyset$. If $p \in J$ then $I \cap J \ne \emptyset$. The only remaining case is that $p=q \not\in J$ in which case $I = \{p\}=\{q\}$ and $q = \sup(I) =\inf(J) \in I \cup J$.
Henceforth we may assume $$\inf(I) < \inf(J) $$ Suppose that $I \cup J$ is not an interval, so there exists $x < z < y$ such that $x,y \in I \cup J$ but $z \not\in I \cup J$. If $x,y$ are both in $I$ this contradicts that $I$ is an interval, and similarly if $x,y$ are both in $J$. So one of $x,y$ is in $I$ and the other is in $J$. Since $\inf(I) < \inf(J)$ and $x < y$, it must be that $x \in I$ and $y \in J$. Since $I$ is an interval with $x \in I$ and $z \not\in I$, it follows that $\sup(I) \le z$. Similarly, $z \le \inf(J)$.
Consider the two inequalities $\sup(I) \le z \le \inf(J)$. If either of these inequalities is strict then $I \cap J = \emptyset$ and $\sup(I) \ne \inf(J)$. Otherwise, if $\sup(I) = z = \inf(J)$, then using that $z \not\in I \cup J$ it follows that $I \cap J = \emptyset$ and that $\sup(I)=\inf(J) \not\in I \cup J$.