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If I understand correctly, most definitions of 'limits' require that the function either a) be defined in an open neighborhood around the relevant point or b) more permissively, that the relevant point is a limit point; the definition of 'continuity' is then given a special case so that functions are continuous at isolated points. Why not extend the notion of 'limit' so that the limit of a function at an isolated point is just whatever the function's value is there? Is there some good reason not to?

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Functions are always defined in open neighborhoods around a point, if "open neighborhood" is defined correctly (relative to the correct subspace). – Qiaochu Yuan Mar 16 '11 at 17:20
What about discontinuous functions? Taking the step function, the one-sided limit is not equal to the function's value at that point. – Blender Mar 16 '11 at 22:31
up vote 10 down vote accepted

You could make that definition I suppose, but what use would it have, and how would it relate to the usual notion of limit?

Let's look at what a limit of a function $f$ at a point $x$ should mean (let's say for a real-valued function on a metric space). You want the limit of $f$ at $x$ to be a real number $L$ such that for all $\varepsilon>0$ there exists a $\delta>0$ such that $0<d(x,y)<\delta$ implies that $|f(y)-L|<\varepsilon$. Now if $x$ is an isolated point, then you could take any $L$ you want, because whatever $\varepsilon$ is, you can choose $\delta$ sufficiently small so that $|f(y)-L|<\varepsilon$ is vacuously true for all $y$ with $0<d(x,y)<\delta$; that is, just make sure that there are no $y$ satisfying the latter condition. For non-isolated points, limits are unique. For isolated points, trying to extend the usual definition leads to making every real number a limit.

For one of the applications of limits, namely continuity, not defining the limit at an isolated point causes no problem. You can say that $f$ is continuous at a point $x$ in the domain if for all $\varepsilon>0$ there exists a $\delta>0$ such that $d(x,y)<\delta$ implies $|f(y)-f(x)|<\varepsilon$. If $x$ is an isolated point, then this will always be true, because for sufficiently small $\delta$ the only $y$ with $d(x,y)<\delta$ is $x$.

Added: I was writing when Alex posted, and part of my post makes a similar point to his. Qiaochu's comment on Alex's post gives an answer to my question at the beginning of my post. Making this definition allows continuity to be defined in terms of respecting limits without making isolated points a special case, something I had overlooked.

Nonetheless, continuity can be defined in terms of respecting limits without actually defining the limit of a function at a point. A function $f$ between metric spaces [resp. topological spaces] is continuous at $x$ if for every sequence [resp. net] $(x_n)_n$ in the domain converging to $x$, $\lim_n f(x_n)=f(x)$. In case $x$ is an isolated point, a sequence converging to $x$ is eventually constantly equal to $x$, so this will be satisfied.

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Maybe I'm missing something, but I don't agree: if there is no point $y$ sucht that $0<d(x,y)<\delta$ for your $\delta$ in the domain of $f$, then $f(y)$ simply does not exist: $f$ is not defined at $y$. Hence the inequality $\mid f(y) - L \mid < \varepsilon$ is not vacuously true: there is no such inequality. Am I wrong? – a.r. Mar 16 '11 at 19:28
Maybe this is more a problem with my wording. If no such $y$ exists, then a statement that involves such $y$ is vacuously true. Whether or not a hypothetical $f(y)$ even makes sense is not considered in that case. As usual with vacuous truth, I can consider what the negation of the statement would be and use the law of excluded middle: to negate, there would have to exist a $y$ such that $0<d(x,y)<\delta$ and $|f(y)-L|\geq \varepsilon$, but no such $y$ exists. But if I am wrong, please inform me! – Jonas Meyer Mar 16 '11 at 19:39
I would have always thought that you cannot put a term that doesn't exist, like that $f(y)$, into an equation... :-? Moreover, to talk about the limit of a function at an isolated point -though maybe not extremely useful in elementary calculus-, makes perfectly sense: it's just the value of the function at this point. And this is not a definition, but a consequence of Rudin's definition -deleting the condition that $x$ must be a limit point. – a.r. Mar 16 '11 at 20:06
@Agustí: What definition exactly do you consider this to be a consequence of? I don't have a particular text reference on hand to dispute this claim. – Jonas Meyer Mar 16 '11 at 20:16
@FreshAir: Yes if the domain is first countable, e.g., a metric space, e.g., a subspace of $\mathbb R^n$, and if $a$ is not an isolated point. – Jonas Meyer Aug 3 '15 at 1:22

Okay, now that I've checked Rudin to see that this really is the definition he gives of a limit of a function, I have an answer, but I don't like it. The motivation behind the definition of $\lim_{x \to a} f(x)$ is that you want to understand what $f$ is doing in a neighborhood of $a$ in order to compare it to what is happening at at $a$. If $a$ is an isolated point, there's nothing to compare.

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+1: I don't really like it either. Fortunately there is no such ambiguity for limits of sequences or nets, and when the notion of limits of functions is introduced in calculus, it is usually for functions defined in a neighborhood (or punctured neighborhood) about a point in $\mathbb{R}^n$. I think this is why I had overlooked the point you made in your first comment on Alex's answer. – Jonas Meyer Mar 16 '11 at 18:14
@Jonas: yes, what really annoys me is that this definition is not equivalent to "the common value of $\lim_{n \to \infty} f(a_n)$ where $a_n \to a$." – Qiaochu Yuan Mar 16 '11 at 18:53
Another possible complication with isolated points is that limits will always exist and always be non-unique. This is not terrible if you are used to limits in arbitrary topological spaces, but for those used to Hausdorff spaces, it's a bit of an issue... – Arturo Magidin Mar 16 '11 at 19:25

The entire point of the notion of a 'limit' is to capture the behavior of a function as it gets "close" to a point, which is not possible for isolated points, thus there is no utility to extending the notion in the way you described. Your definition would make "limit" the same as "value" at these points, which is not very useful.

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Why isn't it useful? If you want the theorem "a function (say, between metric spaces) is continuous if and only if it respects limits" this won't make sense if the domain has isolated points unless you define what limits at an isolated point mean. – Qiaochu Yuan Mar 16 '11 at 17:30
But that would only save you the trouble of saying "except at isolated points", while invalidating numerous theorems about limits such as "a function is continuous if and only if its value is equal to its limit at every point". – Alex Becker Mar 16 '11 at 17:34
I don't understand your second point. Functions are always continuous at isolated points. – Qiaochu Yuan Mar 16 '11 at 17:42
@Qiaochu: Yes, you're right. Sorry, I just whipped that out of my hat without thinking. But I feel it would invalidate statements about when limits can be evaluated, at least as they are presently phrased. – Alex Becker Mar 16 '11 at 17:49
Aren't these two statements the same? 1. "a function (say, between metric spaces) is continuous if and only if it respects limits" 2. "a function is continuous if and only if its value is equal to its limit at every point" – FreshAir Jul 30 '15 at 21:05

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