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If showing an interval $[a,b]$, are $a$ and $b$ implied to be such that $a,b\in\mathbb{R}$ and $a\le b$, by simply writing that, or must they be specified as such?

rubik
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Yaniv Proselkov
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  • I'd even say the strict inequality is implied. – hjhjhj57 Jan 15 '15 at 22:43
  • Yes, $a\le b$ is assumed. Some authors may further assume $a<b$, though most theorems remain true also for intervals of the form $[a,a]$. – Ittay Weiss Jan 15 '15 at 22:45
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    $[a,b] = {x\in\Bbb R\mid a\leq x \leq b}$; in particular if $a>b$ then $[a,b]=\varnothing$; if $a=b$ then $[a,b]={a}={b}$. – Myself Jan 15 '15 at 22:45
  • Well, not necessarily... It's anyway good to precise it ! For example $[2,0]$ wouldn't be shocking but it's not false at all. But usually, you are right :-) – idm Jan 15 '15 at 23:04
  • I've thought about this before, and it seems to me that it varies depending on the context. The definition certainly allows $a>b$ or even (more troublesome) $a=b$. But some theorems about closed sets and closed intervals turn out to be false if you allow empty intervals or (more troublesome) singleton intervals, and people generally don't bother to even mention that the intervals in question must be nontrivial. So I think the answer to this question, like so many others, is, “it depends”. – MJD Jan 15 '15 at 23:21
  • I don't like notations for expressions that tack extra conditions on the theorem statement, because it makes no sense when you try to formalize it. This is the same as $A\sqcup B$ for the "disjoint union" which is the same as a union but tacks an extra assumption $A\cap B=\emptyset$ onto the theorem. There is no formal expression you can write that has this behavior--expressions are like functions that have to return a value, for whatever you give them. All your assumptions should be stated at the beginning, before any expressions that require them - anything else is confusing tricky notation. – Mario Carneiro Jan 15 '15 at 23:39
  • An example: $p\vee\neg p$ is true for any proposition $p$, but if $[a,b]$ tacks an assumption on to propositions containing it then $1\in[a,b]\vee 1\notin[a,b]$ is only true for $b\ge a$, so it follows that $1\in[a,b]$ is not a propositional statement. – Mario Carneiro Jan 15 '15 at 23:44
  • I thought of an example. the famous nested interval theorem says: Let $C_i$ be a family of closed intervals with $C_0 \supset C_1 \supset \cdots$. Then $\bigcap C_i$ is nonempty. Which is true, unless you admit the possibility that $C_i = [a,b]$ with $a>b$, but hardly anyone ever bothers to mention this exception, and for good reason. – MJD Jan 15 '15 at 23:44
  • @MJD Are those proper subsets? If so then each $C_i$ is nonempty, so $a>b$. – Mario Carneiro Jan 15 '15 at 23:45

5 Answers5

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This is defined in the International Standard ISO 31-11.

Snippet extracted from this standard:

$[a, b]$ closed interval in $\Bbb R$ from $a$ (included) to $b$ (included) $$[a, b] = \{x \in\Bbb R ∣ a\le x \le b\}.$$

In other words, there is no a priori assumption on the relationship between $a$ and $b$.

Adam Hughes
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Mufasa
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  • That doesn't seem address the question. ${x:a\leq x\leq b}$ is perfectly sensible when $a>b$, as the comment from Myself mentions. Are you implying that the lack of a specification $a\leq b$ means that no such relationship is assumed? – Kevin Carlson Jan 15 '15 at 22:56
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    @KevinCarlson I think one could argue it does, just not directly. Certainly that unambiguous definition of the set implies the same things as the Myself comment, one should just, perhaps, add the words, "i.e. there is no presupposition on the relationship between $a$ and $b$." – Adam Hughes Jan 15 '15 at 22:57
  • @KevinCarlson - I appologise, I should have made the point clearer. – Mufasa Jan 15 '15 at 23:01
  • @AdamHughes - thank you for clarifying my response. – Mufasa Jan 15 '15 at 23:04
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Most generally, we define intervals on any (partially) ordered set as: $$[a,b]=\{x:a\leq x \leq b\}$$ which implies that if $a>b$, the interval is empty. However, even with this definition, it is perfectly sensical to write $[1,0]$ - it's just that there's nothing in that interval.

I would generally consider that a clause like:

For $a>b$, consider the interval $[a,b]$...

to be a perfectly natural thing to write if you are going to later use the fact that $a>b$ (If you never use this fact, you might as well omit it). The understanding to convey is that we are considering a closed interval, and we are taking $b$ to be its maximum.

It is certainly worth noting that it is not unheard of to think of $[a,b]$ as equal to $[b,a]$, since theorems often are agnostic to the order of the endpoints and it is sometimes more convenient to do so, however, if you do use intervals this way, it is wise to make special note of it. (Likewise though, it is not terribly common to exploit the fact that $[a,b]=\emptyset$ for $a>b$, so if you're going to rely on this in some non-trivial way, it's good to be explicit about that too)

Milo Brandt
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$[a,b]$ is defined as $\{x \in \mathbb{R} ∣ a \leq x \leq b\}$, and it means $a$ and $b$ can also be equal, in that case $[a,b]$ reduces to just a point.

rubik
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rose
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If $a\lt b$, then

$$ [a, b] =\left\{x: x\in\mathbb{R}\land a\leq x \leq b\right\} $$ $$ [a, b) = \left\{ x: x\in\mathbb{R}\land a\leq x \lt b \right\} $$ $$ (a, b] = \left\{ x: x\in\mathbb{R}\land a \lt x \leq b \right\} $$ $$ (a, b)= \left\{ x: x\in\mathbb{R}\land a \lt x \lt b \right\} $$

If $a= b$, then

$$ [a, b] =\{a\}=\{ b \}$$ $$ [a, b) = (a, b]=(a, b)=\varnothing $$

If $a\gt b$, then

$$ [a, b]=[a, b) = (a, b]=(a, b)=\varnothing $$

So no, the interval of $[a, b]$ does not imply that $a\leq b$. We must first know the values of $a$ and $b$ in order to determine what values are in $[a, b]$, if any exist.

k170
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    Your expressions for $a<b$ make no sense, since $\forall x\in \Bbb R,a\le x\le b$ is always false. – Mario Carneiro Jan 15 '15 at 23:33
  • @MarioCarneiro, thank you for this correction. – k170 Jan 16 '15 at 00:30
  • Now you're in good shape. I'll redact my comments and downvote. (Rather, I'll redact my downvote if you make a trivial edit so that I'm permitted to do so.) – Ian Jan 16 '15 at 00:46
  • It doesn't need to be anything of significance, I'm just not allowed to change my vote after this long unless the post was edited in the meantime. (Sorry that I apparently posted my vote without checking for your edit.) – Ian Jan 16 '15 at 00:58
  • The arrows $\implies$ are being used in a very peculiar fashion. I'd rather you use equality signs. – Pedro Jan 16 '15 at 05:44
  • Also, if $a = b$, then $[a,b] = {a} = {b}$, i.e. the set containing $a$ and the set containing $b$. – JimmyK4542 Jan 16 '15 at 05:46
  • @PedroTamaroff, thanks a lot for the correction. – k170 Jan 16 '15 at 05:50
  • @JimmyK4542, you are correct. I've edited my post. – k170 Jan 16 '15 at 05:50
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Yes, it is implicitly understood that $a<b$ since the first number always represents the left hand endpoint, and the second number the right hand endpoint.

Tim Raczkowski
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