Does the interval notation $[a,b]$ imply that $a<b$?

Question

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?

Answer 1

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.

Answer 2

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 $b$.

Answer 3

$[a,b]$ 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.

Answer 4

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.

Answer 5

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)