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Are $(\infty,\infty)$ and $(-\infty,-\infty)$ intervals?

Also, is (1,1) an interval? if so, is it just an empty interval?

sucksatmath
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3 Answers3

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Interval notation is exactly that: notation

It is a shorthand way of denoting certain types of subsets of $\mathbb{R}$.

$(a,\infty)$ for example is just an abbreviation for $\{x\in\mathbb{R}\,\vert\,x>a\}$

$(a,b)$ is shorthand for $\{x\in\mathbb{R}\,\vert\,a<x<b\}$

Given this understanding of the notation, constructions such as you ask about are not intervals. If you wish to call $(1,1)$ the empty set that is not particularly objectionable, but to call it an interval would be non-standard.

John Wayland Bales
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  • @user21820 Yes, I agree that the answer can depend upon one's definition of interval. I prefer the following definition: An interval is a collection of real numbers (note the plural form) with the property that if $a$ and $b$ are in the set and $a<b$ and $x$ is a number satisfying $a<x<b$ then $x$ is also in the set. By this definition an interval must contain two numbers. Perhaps there are situations where it is useful to consider the empty set to be an interval. In such a situation one would define the term differently. https://en.wikipedia.org/wiki/Interval_(mathematics) – John Wayland Bales Jan 20 '17 at 08:12
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    In mathematics, "a collection of real numbers" can be an empty collection, and also an empty-set vacuously satisfies the condition because you cannot find members $a,b$ such that $a<b$. So I don't agree with the phrases "not intervals" and "abuse of notation". If you change to "not standard interval notation" and "uncommon use of notation" it would be much less disputable. – user21820 Jan 20 '17 at 09:07
  • @user21820 Rather than turn this into a discussion of the philosophy of mathematics I will do as you ask. Note: I thought your answer was perfectly legitimate and am not one of those who downvoted it. – John Wayland Bales Jan 20 '17 at 14:58
  • Thanks! Do you want me to delete my comments? By the way I never suspected you as a downvoter. – user21820 Jan 20 '17 at 15:00
  • I was not offended by your comments so I do not care whether or not you delete them. – John Wayland Bales Jan 20 '17 at 15:02
  • Oh okay; I just thought they are no longer relevant since you've edited your answer. – user21820 Jan 20 '17 at 15:02
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At least according to the standard definition given on Wikipedia, which is that an interval of reals is a convex set of reals (every real between any two members is also an member), $(1,1)$ is indeed the empty set and is an interval by definition.

$(\infty,\infty)$ can be and is readily interpreted in the affinely-extended reals in the same manner, giving an empty set as well. Same for $(-\infty,-\infty)$. However, notation wise they do not make sense if we insist on restricting to the real line alone.

[Two people now have downvoted my answer even though its first paragraph says essentially the same thing as Asaf Karagila's answer. Please read his answer properly as it explains quite well the basic logic needed to understand this "vacuous truth". Also, most people adopt such a definition that does not exclude empty intervals because it leads to more elegant theorems (see Noah Schweber's answer).]

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user21820
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    It seems a bit excessive to directly call out other users as incorrect in your answer, given that this is more a dispute over the correct definition rather than an error in anyone's answer. Wikipedia does not have the single, authoritative definition, and many people choose to use other ones. This is something that many users of this website seem to miss. – Vik78 Jan 20 '17 at 05:20
  • @Vik78: It's a mainstream definition. We can't keep arguing over definitions when asked a simple question. I didn't even say they are wrong under other definitions, nor did I downvote John's answer, so it's excessive for you to downvote my answer. – user21820 Jan 20 '17 at 05:35
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    I didn't downvote your answer. I still think it'd be more professional to remove their names from your answer, but do whatever you want. – Vik78 Jan 20 '17 at 05:47
  • @Vik78: Oh I thought you did because the downvote was at the same time as your comment. For you I'll remove the reference, but I don't see any necessity to since I was merely stating a matter of fact. – user21820 Jan 20 '17 at 05:50
  • It's hardly "contrary" to anything. If you're using different definitions, you're using different definitions. You're not talking about the same thing. Thus, by definition, your results could not contradict someone else's who is talking about something else. I think the phrase "contrary to other existing answers" is unnecessarily argumentative. Otherwise this is a great answer. +1, -1, thus I didn't vote. :) – Wildcard Jan 20 '17 at 06:06
  • @Wildcard: It's contrary because the other answers fall by exactly the same criterion as Vik78 claimed to espouse. They give a categorical "no" or "abuse of notation", whereas my answer is the far more precise and accurate one. I don't know what you expect me to say instead without leaving out the fact that the other answers are not valid according to a highly mainstream definition. – user21820 Jan 20 '17 at 06:09
  • What is a "standard definition" on Wikipedia? Something may be or not may be an interval according the (current) definition on Wikipedia. But Wikipewdia does not define standards. – miracle173 Jan 20 '17 at 15:09
  • @miracle173: I just meant "a standard definition, specifically the one on Wikipedia". If you think my phrasing is ambiguous I'll change it. – user21820 Jan 20 '17 at 15:11
  • At the risk of starting this up again perhaps I should explain why I originally said "abuse of notation" (which I do not consider necessarily a bad thing--we mathematicians abuse notation all the time). While the empty set is a subset of every set, it is not an element of every set. So while the empty set is a subset of the set of all intervals, that does not make the empty set an interval. So while I agree that the only consistent meaning that could be given to $(1,1)$ is the empty set, that does not make the empty set an interval, so it could be considered a misuse of the notation. – John Wayland Bales Jan 20 '17 at 19:56
  • Here is another contrary view: http://math.stackexchange.com/questions/1228307/why-is-the-empty-set-considered-an-interval – John Wayland Bales Jan 20 '17 at 20:01
  • @JohnWaylandBales: Based on the Wikipedia definition (which is used in many textbooks), you're totally wrong about that. All the conditions are satisfied and have nothing to do with the empty-set being a subset of the set of intervals. And Asaf's answer at the post you linked says exactly what I say, contrary to your claim. So I don't understand the downvote. – user21820 Jan 21 '17 at 01:01
  • @user21820 I upvoted your answer. And I deliberately linked to an opinion in opposition to my own to give a balanced view. When I said a contrary opinion I meant contrary to my opinion. – John Wayland Bales Jan 21 '17 at 02:08
  • @JohnWaylandBales: Well I wasn't sure if it was you, and so thanks for mentioning the link, but my point about your wrong reasoning in your comment still stands. – user21820 Jan 21 '17 at 02:23
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    I also sought other opinions by posing a question here where you might want to weigh with your view (which is the majority view) http://math.stackexchange.com/questions/2106574/must-real-number-interval-be-defined-in-such-a-way-as-to-include-the-empty-set – John Wayland Bales Jan 21 '17 at 02:29
  • @JohnWaylandBales: Yes I saw that question and linked to it in my most recent edit. I didn't feel a need to say anything there since you already had accepted an answer that I agreed with, but since you suggested I'll add another one emphasizing the key issues. – user21820 Jan 21 '17 at 02:34
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To add to other answers:

$(a,b) = \{x \in \mathbb{R}: a < x < b\}$

$(a,b] = \{x \in \mathbb{R}: a < x \le b\}$

$[a,b) = \{x \in \mathbb{R}: a \le x < b\}$

$[a,b] = \{x \in \mathbb{R}: a \le x \le b\}$

And so forth. Similarly, $\infty$ or $-\infty$ as interval bounds are simply notation for no bounds on that end of the interval.

Also, an interval with one or no elements like $[a,a]$ or $(a,a)$ are often called degenerate intervals if you want to learn more about them.

Vedvart1
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