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I was thinking about the epsilon-delta definition of a limit and began to wonder if it would work with closed intervals instead of open intervals. That is, for some function $f$, would the folowing definition work? $$\forall \epsilon>0\,\exists\delta>0\,(0<\lvert x-c\rvert≤\delta \, \implies \lvert f(x)-L\rvert≤\epsilon) $$ I've been trying to think of a counterexample, but can't seem to find one.

edit: If it does work, why choose open intervals rather than closed?

Simplicio
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    It should work fine, but be careful of the possible choice of $x$. – edm Aug 27 '17 at 02:32
  • In fact, in more advanced course, for example "Mathematical Analysis", the definition of the limit doesn't require any open interval, but rather a set $E$ with the restriction: $c$ is an adherent point(also called cluster point, limit point) of $E$. OK, I write the full version for you:

    Let $E$ be a set, $f:E\to\Bbb R$, $c$ is an adherent point of $E$ and $L\in\Bbb R$. If $\forall \epsilon>0,~\exists\delta >0,~\forall x\in E,(0<|x-c|<\delta\to |f(x)-L|<\epsilon)$, then we called the limit of $f$ at $c$ is $L$.

     – Eric Aug 27 '17 at 02:48
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    The two statements are equivalent. If it's true for an open or closed neighborhood then it is also true for the nested closed and open neighborhoods within. I think we use open neighborhoods because they are more general-- if we choose a point that is directly "on the edge" it's not as general as a point that is "somewhere inside". Also we use the concept of delta epsilon for other things than limits, example, definition of open where less than is just better habit. – fleablood Aug 27 '17 at 03:17

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