Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have below a beginning of a theorem:

If a function $f:I \rightarrow \mathbb{C}$ defined on an interval $I$ of length $p$ can be expanded to a piecewise differentiable function on $\overline{I}$, then will...

What does $\overline{I}$ mean in this context?

share|cite|improve this question
usually the closure of $I$. That is, if $I$ is any of $(a,b), [a,b), (a,b], [a,b]$, then $\overline{I}= [a,b]$. – user20266 Jun 6 '12 at 19:45
Thank you sir, I will leave it as it is for now. – Paul Slevin Jun 6 '12 at 19:50
up vote 3 down vote accepted

Here $\overline I$ means the closure of $I$ - in general this is the smallest closed set which contains $I$. If $I \subseteq \mathbb{R}$ is an interval, then it is just the interval with the endpoints included.

For a more general example, in $\mathbb{R}^2$ you have the set $B = \{ (x,y) \in \mathbb{R}^2 : | (x,y) | < 1 \}$, the open ball of radius $1$, centred at the origin. If we take its closure we get $\overline B =\{(x,y) \in \mathbb{R}^2 : |(x,y)| \le 1 \}$, the open disc which contains the boundary. This is the two-dimensional analogue of an interval.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.