3

For example, $\varepsilon$, $\delta$ definition is based on the open ball about the point a with which we take the limit respect to.

So my question is what is the problem with having a close ball about the point a?

I assume it has something to do with having both the limit $\lim{x \to a^-}$ and $\lim{x \to a^+}$.

This is my first time asking. I apologize if there exists any format mistakes in my question.

ArdaSenyurek
  • 41
  • 1
    You could use closed balls as long as they contain an open one to avoid the closed ball ${a}$. – copper.hat Jun 17 '22 at 05:35
  • It doesn't make any difference as long as $\mathbf{a}$ is an interior point of the closed ball. In fact, you can use any neighborhood of $\mathbf{a}$. A set $B$ is called a neighborhood of $\mathbf{a}$ if $\mathbf{a}$ is an interior point of $B$. A closed ball centered at $\mathbf{a}$ is clearly a neighborhood of $\mathbf{a}$. – Onur Oktay Jun 17 '22 at 05:36
  • In many theorems you can indeed use closed balls, or indeed any collection of subsets containing an open ball around $a$. This in most cases does not make theorems either stronger or weaker. (As said in other comments, this excludes the closed ball with radius $0$.) In fact, there is a generalisation: many theorems in Calculus also work when the set of sets you want to consider is a filter of subsets (a nonempty collection closed for finite intersections and for supersets) - e.g. supersets of intervals $(a, a+\epsilon)$ if you want to study $\lim_{x\to a+}$, for example. –  Jun 17 '22 at 05:39
  • Similar: https://math.stackexchange.com/questions/264148/why-does-the-definition-of-limits-of-a-function-have-strict-inequality – Hans Lundmark Jun 17 '22 at 06:15
  • Some time ago, some authors defined a neighbourhood of $x$ to be any set that contained an open set containing $x$, that is, the neighbourhood itself did not need to be open. I am not sure what this added to the overall scheme of things, but it certainly threw me when I was learning. – copper.hat Jun 17 '22 at 17:52
  • @copper.hat - what it adds is the ability to say such things as "compact neighborhood of $x$" instead of "compact set containing a neighborhood of $x$". After saying that or similiar phrases a few times, the advantages of not requiring neighborhoods to be open becomes apparent. – Paul Sinclair Jun 18 '22 at 00:16
  • @PaulSinclair Makes sense. It confused me the first time I encountered it. – copper.hat Jun 18 '22 at 03:01

0 Answers0