2
$\begingroup$

We did limits of functions recently and I am wondering why we always required that $x_0$ is a cluster point of the domain. Why would taking the limit not work if $x_0$ is not a cluster point?

Our definition of a limit of a function is

Let $D \subseteq \mathbb R$ be a subset, $x_0$ a cluster point of $D$ and $f: D \to \mathbb R$ a function. We say $f$ converges to $L \in \mathbb R$ and write $\lim_{x \to x_0} f(x) = L$ $\iff \forall \varepsilon \gt 0 \, \exists \delta \gt 0 \, \forall x \in D \setminus \{x_0\}: |x-x_0| \lt \delta \implies |f(x)-L| \lt \varepsilon$

Our definition of a cluster point is

Let $D \subseteq \mathbb R$ be a subset and $x_0\in \mathbb {R}$. We say $x_0$ is a cluster point of $D$ $\iff$ for every $\delta \gt 0$ we have $D \cap (x_0-\delta, x_0-\delta) \setminus \{x_0\} \neq \emptyset$

$\endgroup$
  • $\begingroup$ Can you cook up an example of an $x_0$ that’s not a cluster point, and see what the limit definition boils down to without that provision? $\endgroup$ – Lubin Jan 4 '18 at 19:48
  • 1
    $\begingroup$ @Lubin Do you mean if $x_0$ is not a cluster point the statement could be vacuously true because for the $\forall x \ldots$ there might not actually be such $x$? $\endgroup$ – philmcole Jan 4 '18 at 19:51
  • 1
    $\begingroup$ Just so: since there won’t be any such $x$’s, the desired inequality $\vert f(x)-L\vert<\varepsilon$ will always be true. Independent of $L$ ! So every $L\in\Bbb R$ will be a limit, in other words, the limit is not unique. $\endgroup$ – Lubin Jan 4 '18 at 19:55
  • 1
    $\begingroup$ The concept of limit involves dealing with behavior of a function in a neighborhood of a given point and by definition requires that the neighborhood contains infinite number of points. This fact is not stressed in many textbooks but Hardy clearly mentions this in his classic text A Course of Pure Mathematics before introducing the concept of limits. $\endgroup$ – Paramanand Singh Jan 5 '18 at 8:02
  • $\begingroup$ Related: math.stackexchange.com/questions/1815425/… $\endgroup$ – philmcole Jan 5 '18 at 10:49
0
$\begingroup$

You need that $x_0$ is a cluster point in order to find $\forall \varepsilon \gt 0$, $x$ such that

$$|x-x_0| \lt \delta$$

otherwise you can't define a limit.

$\endgroup$
  • $\begingroup$ Thanks! But why exactly is this necessary? The definition of continuity for example is "$f: D \to \mathbb R$ is continuous at $x_0 \in D$ iff $\forall \varepsilon \gt 0 \, \exists \delta \gt 0 \, \forall x \in |x-x_0| \lt \delta \implies |f(x)-f(x_0)| \lt \varepsilon$". There $x_0$ doesn't need to be a cluster point, so if $x_0$ is a single isolated point in $D$ then the statement is always true. $\endgroup$ – philmcole Jan 4 '18 at 19:59
  • $\begingroup$ @philmcole This is simply a necessary ingredient for the definition. Since we use limits to calculate derivatives (e.g. tangent line at a point) I guess limits defined in others ways should be useless without this requirment on $x_0$. $\endgroup$ – user Jan 4 '18 at 20:04
  • 2
    $\begingroup$ @philmcole: You can think of it like this, intuitively: The limit at $x_0$ is what the surrounding points think that the value $f(x_0)$ should be. If there is a limit, it means that all the neighbors agree on this, and then $f(x_0)$ has to take that value in order for $f$ to be continuous at $x_0$. But if there are no surrounding points (i.e., if $x_0$ is an isolated point in the domain of $f$), then there is no opinion to obey, and $f(x_0)$ is free to take on any value without upsetting the neighbors, and we also consider $f$ to be continuous at $x_0$ in this case. $\endgroup$ – Hans Lundmark Jan 4 '18 at 20:22
  • 3
    $\begingroup$ Your answer is incorrect. Remove the conditon "$x_0$ is a cluster point" and you still have a valid definition. However, any real function on $D$ would then satisfy the definition vacuously. Since such a definition would have zero value, we wish to avoid it. That's why the conditon "$x_0$ is a cluster point" is part of the definition. $\endgroup$ – zhw. Jan 5 '18 at 0:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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