# Why is the empty set considered an interval?

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

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 $$a, we have that $$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.

• Re. "Nowhere in books ...." In Munkres Topology 2nd edition p.84 he defines intervals in $A$ based on $a, b \in A$ with $a \lt b$. – Tom Collinge Jun 15 '17 at 11:59
• 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 – Asaf Karagila Jun 15 '17 at 12:27
• According to your definition $\mathbb R$ is an interval, but it has no endpoints. – bof Aug 14 '17 at 6:37
• @bof: That's a really weird way of saying "I think that you meant bounded intervals have endpoints". – Asaf Karagila Aug 14 '17 at 6:38

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.

Why is "for all $x\in\varnothing$, $P(x)$" true, but "there exists $x\in\varnothing$ such that $P(x)$" false?
• What is $\;P\;$ in your answer? – Timbuc Apr 10 '15 at 9:50
• @Timbuc You just "match" the second line with the third: $P(x)$ is $\forall y\in I, \forall z\in I, x<z<y \implies z\in I$. – Stop hurting Monica Apr 10 '15 at 9:57