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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?

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    I think the answer is no. – 2017-01-10
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    [Wikipedia](https://en.wikipedia.org/wiki/Interval_(mathematics)) anwers all of these questions. – 2017-01-10
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    See https://proofwiki.org/wiki/Definition:Real_Interval – 2017-01-10
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    See [extended reals](https://en.wikipedia.org/wiki/Extended_real_number_line) – 2017-01-10

3 Answers 3

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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

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.

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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 $a$x$ is a number satisfying $a$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) – 2017-01-20
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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 – 2017-01-20
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    @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. – 2017-01-20
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    Thanks! Do you want me to delete my comments? By the way I never suspected you as a downvoter. – 2017-01-20
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    I was not offended by your comments so I do not care whether or not you delete them. – 2017-01-20
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    Oh okay; I just thought they are no longer relevant since you've edited your answer. – 2017-01-20
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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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    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. – 2017-01-20
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    @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. – 2017-01-20
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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. – 2017-01-20
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    @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. – 2017-01-20
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    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. :) – 2017-01-20
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    @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. – 2017-01-20
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    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. – 2017-01-20
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    @miracle173: I just meant "a standard definition, specifically the one on Wikipedia". If you think my phrasing is ambiguous I'll change it. – 2017-01-20
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    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. – 2017-01-20
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    Here is another contrary view: http://math.stackexchange.com/questions/1228307/why-is-the-empty-set-considered-an-interval – 2017-01-20
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    @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. – 2017-01-21
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    @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. – 2017-01-21
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    @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. – 2017-01-21
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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 – 2017-01-21
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    @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. – 2017-01-21
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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.