I have a question on union of intervals.

My teacher says the union of intervals is not an interval.

Is this always true?

I mean

$[0,2] \cup [4,5]$ is not an interval, because $[0,2] \cap [4,5]= \varnothing ,$

but what about

$ (0,8) \cup (7,9).$

I think this is an interval because $(0,8) \cup (7,9)=(0,9)$.

  • 1
    $\begingroup$ Note that $[0,2]\cap(2,3]=[0,3]$ is an interval, but $[0,2]\cap(2,3]=\emptyset$. Therefore, it is not enough to use that the intersection is empty to guarantee that the union is not an interval. In your first case, the problem is that $3\not\in[0,2]\cup[4,5]$, but $2\leq 3\leq 4$ and $2$ and $4$ are in the union. $\endgroup$ Jan 10, 2017 at 23:09

4 Answers 4


For a statement to be true in mathematics, it must always be true without exception. The statement "The union of two intervals is an interval" is false because even though, sometimes, the union of two intervals is an interval (as in your second example), there are counterexamples. Since the statement is not always true (as in your first example), we say that it is false.

  • 8
    $\begingroup$ To expand a little, usually in mathematics we understand "The union of two intervals is an interval" to be a general statement along the lines of "for all X, for all Y, X is an interval implies (Y is an interval implies X ∪ Y is an interval)". Which of course is false. In some contexts we might understand it to be a specific statement about two particular intervals, which could be true or false depending on the intervals chosen, or could be the statement of a constraint. For example: "The union of two open intervals is an interval. Prove that their intersection is not empty." $\endgroup$ Jan 11, 2017 at 0:46
  • 3
    $\begingroup$ @SteveJessop I think that should be an answer. $\endgroup$
    – jpmc26
    Jan 11, 2017 at 2:58
  • $\begingroup$ Thank you very much. $\endgroup$
    – Sebastiano
    Jul 26, 2020 at 23:07

The union of intervals is not always an interval. Sometimes it is, sometimes it isn't. Your second example is in fact an interval, while your first example is not.


The problem is that you and your teacher have different interpretations of an ambiguous English statement as a precise mathematical statement.

Your teacher's intended meaning.

"The union of [any two] intervals is not [necessarily] an interval". This statement is easily proved true by providing two disjoint intervals whose union is not an interval, such as [0,2] and [4,5].

In other words, there is counterexample to the false statement "For all intervals $X$ and $Y$, $X \cup Y$ is also an interval."

Your interpretation

"The union of [two particular] intervals is not [ever] an interval". This statement is easily proved false providing an example of two overlapping intervals whose union is an interval, such as (0,8) and (7,9).


The union of intervals is sometimes, but not always, an interval. So your teacher is right.

  • $\begingroup$ Thanks, but in my second example is an interval, right? $\endgroup$
    – Dima
    Jan 10, 2017 at 22:50
  • $\begingroup$ @Dima Yes, it is. $\endgroup$ Jan 10, 2017 at 22:56
  • 4
    $\begingroup$ The teacher's statement is not right unless you replace "not" with "not necessarily". $\endgroup$ Jan 10, 2017 at 23:45
  • $\begingroup$ @AkivaWeinberger: It's standard in English that "not" can semantically take scope over the entire clause it resides in. Hence, for example, "All that glitters is not gold", meaning "It's not the case that {All that glitters is gold}." So although the teacher's statement can be interpreted in a way that makes it wrong, its intended meaning is correct. $\endgroup$
    – ruakh
    Jan 11, 2017 at 7:51
  • 1
    $\begingroup$ @ruakh Maybe. I've always found the "All that glitters is not gold" sentence to sound unnatural, though. $\endgroup$ Jan 11, 2017 at 12:03

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