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What is the definition of an interval and why is the empty set an interval by that definition?

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Given an ordered set $(A,\leq)$ an interval is a subset which is convex. Namely $I$ is an interval in $A$ if whenever $a,b\in I$ then for every $x$ such that $a<x<b$, we have that $x\in I$ as well.

In the real numbers, because the order is complete, it follows that every bounded interval has endpoints, and may or may not include them. For example $[0,1]$ is the interval of all numbers from $0$ to $1$, including these two, and $(\sqrt2,42]$ is the interval of all numbers strictly larger than $\sqrt2$ but not strictly larger than $42$.

But nowhere in the books it is said that the endpoints of the interval have to be different from one another. What would be $(0,0)$? It would be all the numbers strictly between $0$ and $0$. But no such numbers exist, so $(0,0)=\varnothing$.

And indeed, whenever $a,b\in\varnothing$, and $x$ lies between $a$ and $b$, it follows that $x\in\varnothing$ as well. If you want to claim that this is false, you need to come up with actual $a,b\in\varnothing$ and $x$ between them which is not an element of $\varnothing$. Coming up with $x$ is easy, but coming up with $a$ and $b$ is impossible. So the definition is satisfied vacuously and $\varnothing$ is an interval.

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  • $\begingroup$ Re. "Nowhere in books ...." In Munkres Topology 2nd edition p.84 he defines intervals in $ A$ based on $a, b \in A$ with $a \lt b$. $\endgroup$ – Tom Collinge Jun 15 '17 at 11:59
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    $\begingroup$ No, actually he says that if $a<b$ are two real numbers, then there are subsets determined by $a$ and $b$ called intervals. But if you look at Royden--Fitzpatrick's "Real Analysis" (4th ed.), while they begin with a definition similar to that of Munkres, on p. 15 they also say that an interval can be empty or with a single point. This is exactly the case where $a=b$. I don't know why Munkres defines it this way, because I haven't worked through his book when I studied topology. While I'm sure he has a reason, allowing $a=b$ and degenerate intervals can be very useful. But thanks for the remark $\endgroup$ – Asaf Karagila Jun 15 '17 at 12:27
  • $\begingroup$ According to your definition $\mathbb R$ is an interval, but it has no endpoints. $\endgroup$ – bof Aug 14 '17 at 6:37
  • $\begingroup$ @bof: That's a really weird way of saying "I think that you meant bounded intervals have endpoints". $\endgroup$ – Asaf Karagila Aug 14 '17 at 6:38
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The definition of a real interval is: a subset $I$ of $\Bbb R$ such that

$$\forall x\in I, \forall y\in I,\forall z\in \Bbb R, x<z<y \implies z\in I$$

Notice that $\forall x\in \emptyset, P(x)$ is always true by definition.

See also vacuous truth on Wikipedia, and these questions on MSE:

Why is predicate "all" as in all(SET) true if the SET is empty?

Why is "for all $x\in\varnothing$, $P(x)$" true, but "there exists $x\in\varnothing$ such that $P(x)$" false?

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  • $\begingroup$ What is $\;P\;$ in your answer? $\endgroup$ – Timbuc Apr 10 '15 at 9:50
  • $\begingroup$ @Timbuc : a predicate. $\endgroup$ – Stop hurting Monica Apr 10 '15 at 9:52
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    $\begingroup$ Oh, I know that: which one, though? Until your third line it doesn't appear at all yet you use it in your answer. $\endgroup$ – Timbuc Apr 10 '15 at 9:54
  • $\begingroup$ @Timbuc You just "match" the second line with the third: $P(x)$ is $\forall y\in I, \forall z\in I, x<z<y \implies z\in I$. $\endgroup$ – Stop hurting Monica Apr 10 '15 at 9:57

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