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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?

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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
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up vote 2 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.

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