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Normally, when we're dealing with intervals $[a,b]$, it is at least implied that $a\leq b$. In this case, there are multiple equivalent ways to define what the interval actually is:

  1. $[a,b]=\{x\in\mathbb{R}\colon a\leq x\leq b\}$
  2. $[a,b]:=\operatorname{conv}(\{a,b\})$.

While the former definition is the one usually taught, the latter seems rather more elegant to me, though the convex-hull operator is itself defined by a more complicated set-builder notation. As both are equivalent, the distinction is unnecessary.

It gets interesting when we have $[a,b]$ with $a>b$. Usually, I would consider this simply wrong, but sometimes you have a general interval $[a,b]$ and then it can be tedious and contraproductive to always have to require $b\geq a$. What such an interval is comes out differently from both definitions:

  1. Always the empty set. This is the usual interpretation according to Wikipedia, but it's not necessarily very useful.
  2. The set with "correctly-ordered" borders $[b,a]$.

The latter result seems more useful to me, for instance we have always $$\mu([a,b]) = \mu([b,a])$$ $$ \int\limits_a^b\mathrm{d}x\ f(x) = \int_{[a,b]}\!\!\!\mathrm{d}x\ f(x)\cdot\operatorname{sgn}(b-a) $$ and so on.

What reasons are there to not just always define intervals this way?

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I have seen $[a,b]$ used to denote the oriented line segment connecting $a$ and $b$ in the complex plane. –  Jonas Meyer Aug 14 '11 at 21:41
    
Why do you want to use the other definition for a real interval? You are not satisfied with the simple, standard definition? By defining an interval without looking who is the left and right endpoint you lose the characteristic property of $\Bbb{R}$ compared to $\Bbb{R}^k,\ k \geq 2$ and that is 'order'. In many problems, the order of the endpoints is useful. What do you mean when you say that the second definition is more useful to you? What other uses does it have? –  Beni Bogosel Aug 14 '11 at 21:44
    
@Beni I am ok with the simple, standard definition if it does always require $a\leq b$. But extending this definition to $a>b$ so that such an interval is always the empty set is only useful if we're really interested in $a\leq x$ and $x\leq b$ rather than just "$x$ lies between $a$ and $b$" or "$x$ runs from $a$ to $b$", which seems by far the more important application to me. –  leftaroundabout Aug 14 '11 at 21:59
    
Slightly off-topic perhaps, but in interval arithmetic for auto-validating numerical methods, $[a,b]$ with $a>b$ is sometimes used to represent the set $[a,+\infty) \cup (-\infty,b]$. –  Hans Lundmark Aug 15 '11 at 6:30
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2 Answers

up vote 2 down vote accepted

Because of the usual left to right reading order in European languages, $[a,b]$ has a useful connotation of order. However, your suggestion certainly has merit. If it were the current convention, there would be no strong arguments for change.

However, it is not the current convention. In elementary mathematics, changing the convention would instantly make most books very slightly obsolete. That might make the idea attractive to publishers, but surely not to anyone else.

History has left us with quite a number of unfortunate notations. If one introduces a new notation, or a new interpretation of a notation, ordinarily one of the following happens: (i) The new notation is not widely adopted, and quickly dies or (ii) The new notation or interpretation is attractive to many, and is adopted by many. In case (ii), we end up with two notations or interpretations.

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I can think of two reasonable interpretations. If you think of $[a, b]$ as defining the set $\{ x : a \le x \le b \}$ then clearly this is empty if $a > b$. This would be an appropriate convention in the context of general posets, where sums over all intervals in a poset should be interpreted like this.

The second is to regard $[a, b]$ as being the negative of $[b, a]$. That is, identify $[b, a]$ with its indicator function: then $[a, b]$ should be the negative of the indicator function. This would be an appropriate convention for defining signed integrals over intervals, which is really a special case of a contour integral.

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This is a great solution, the only problem I have with it is that it makes all intervals objects of more complicated kind than just sets. Could these be described as multisets? –  leftaroundabout Aug 14 '11 at 22:01
    
@left: well, functions are more general than multisets. From the perspective of measure theory it's natural to regard a measurable function as a generalized or weighted indicator function. –  Qiaochu Yuan Aug 14 '11 at 22:07
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