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I was sitting in analysis yesterday and, naturally, we took the limit of some expression. It occurred to me that "taking the limit" of some expression abides the rules of a linear transformation $$\lim_{x \rightarrow k}\ c(f(x)+g(x)) = c \lim_{x \rightarrow k} f(x) + c\ \lim_{x \rightarrow k} g(x),$$ and (my group theory is virtually non existent) appears also to be a homomorphism: $$\lim_{x \rightarrow k} (fg)(x) = \lim_{x \rightarrow k} f(x)g(x), $$ etc.

Anyway, my real question is, what mathematical construct is the limit?

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    $\begingroup$ I would avoid defining $\lim$ as a function/relation/etc. but rather $\lim_{x\to k} f(x)=a$ as a notational convention or abbreviation for a much more complicated statement. You might as well ask whether $\dim$ is a function from vector spaces to natural numbers, but you will have trouble defining the "set of all vector spaces". Instead writing $\dim(V)=n$ is an abbreviation of a much longer statement about bases, linear indendence, etc. $\endgroup$
    – M. Winter
    Commented Feb 2, 2020 at 20:13
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    $\begingroup$ This is exactly what I'm wondering. Of course I understand that it is a convention of notation, but it's not only a convention of notation. It can't be "turtles all the way down". There has to be some foundational notion at the bottom of the abstraction pile which lends itself to a thorough understanding of what's happening when one takes a limit. That was my question. @M.Winter $\endgroup$ Commented Feb 3, 2020 at 14:37
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    $\begingroup$ Similar question (old, but it turned up on the front page recently): What kind of "mathematical object" are limits? $\endgroup$ Commented Feb 3, 2020 at 20:26

3 Answers 3

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In general, let $X, Y$ be topological spaces, and $x_0$ a non-isolated point of $X$. Then strictly speaking, "$\lim_{x\to x_0} f(x) = L$" is a relation between functions $f : X \to Y$ and points $L \in Y$ (the equality notation being misleading in general).

Now, if $Y$ is a Hausdorff topological space, it happens that this relation is what is known as a partial function: for any $f : X \to Y$, there is at most one $L \in Y$ such that $\lim_{x\to x_0} f(x) = L$. Now, for any relation $R \subseteq (X \to Y) \times Y$ which is a partial function, we can define a corresponding function $\{ f \in (X \to Y) \mid \exists y \in Y, (f, y) \in R \} \to Y$ by sending $f$ satisfying this condition to the unique $y$ with $(f, y) \in R$. Then that somewhat justifies the "equality" in the notation $\lim_{x\to x_0} f(x) = L$, though you still need to keep in mind that it is a partial function where $\lim_{x\to x_0} f(x)$ is not defined for all $f$. (This part relates to the answer by José Carlos Santos.)

Building on top of this, in the special case of $Y = \mathbb{R}$, we can put a ring structure on $X \to Y$ by pointwise addition, pointwise multiplication, etc. Then $\{ f : X \to \mathbb{R} \mid \exists L \in \mathbb{R}, \lim_{x\to x_0} f(x) = L \}$ turns out to be a subring of $X \to \mathbb{R}$, and the induced function from this subring to $\mathbb{R}$ is a ring homomorphism. (More generally, this will work if $Y$ is a topological ring. Similarly, if $Y$ is a topological vector space, then the set of $f$ with a limit at $x_0$ is a linear subspace of $X \to Y$ and the limit gives a linear transformation; if $Y$ is a topological group, you get a subgroup of $X \to Y$ and a group homomorphism; and so on.)

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    $\begingroup$ +1, though I don't think your hesitation with respect to the equals sign is justified; $\frac{6}{x}$ is not defined when $x = 0$, but this certainly doesn't stop us from writing $\frac{6}{2} = 3$. $\endgroup$
    – ruakh
    Commented Feb 2, 2020 at 1:34
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    $\begingroup$ Perhaps Daniel is thinking along these lines: quantamagazine.org/… $\endgroup$ Commented Feb 2, 2020 at 5:07
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    $\begingroup$ That comment was in the context of the general case (not necessarily Hausdorff $Y$) in which you do not have uniqueness of the limit. (So, roughly the same objection that many people have to the $f = O(g)$ notation.) $\endgroup$ Commented Feb 2, 2020 at 7:00
  • $\begingroup$ You don't need a space to create Lim, but limits on spaces are just mostly known. $\endgroup$
    – sanaris
    Commented Feb 2, 2020 at 15:29
  • $\begingroup$ @sanaris How would you take a limit without a space? $\endgroup$
    – Jam
    Commented Feb 2, 2020 at 21:50
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Let $I\subset\mathbb R$ be a subset of $\mathbb R$ such that $k$ is an accumulation point of $I$. Let$$R=\left\{f\colon I\longrightarrow\mathbb R\,\middle|\,\lim_{x\to k}f(x)\text{ exists}\right\}.$$Then $(R,+,\times)$ is a ring and the map$$\begin{array}{ccc}R&\longrightarrow&\mathbb R\\f&\mapsto&\lim_{x\to k}f(x)\end{array}$$is a ring homomorphism.

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  • $\begingroup$ what if $f$ is unbounded around $k$ or am I missing something? E.g assume $0\in I$, then $\frac{1}{x^2} \xrightarrow[x\to 0]{} \infty$. $\endgroup$
    – AlvinL
    Commented Feb 1, 2020 at 16:49
  • $\begingroup$ Such a function $f$ will not belong to $R$; in the words, the limit $\lim_{x\to k}f(x)$ will not exist then. $\endgroup$ Commented Feb 1, 2020 at 16:54
  • $\begingroup$ Take $I=(0,1)$. Then $f\in R$ and its limit is infinity as $x\to 0$, where $0$ is an accumulation point of $I$. The limit does exist. $\endgroup$
    – AlvinL
    Commented Feb 1, 2020 at 16:55
  • $\begingroup$ No, it doesn't. I was thinking only about the case in which $\lim_{x\to k}f(x)$ exists in $\mathbb R$. $\endgroup$ Commented Feb 1, 2020 at 16:57
  • $\begingroup$ Ah, right, my bad. Would be helpful to say limit exists and is finite or some such. $\endgroup$
    – AlvinL
    Commented Feb 1, 2020 at 16:58
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To be more precise, "taking $\limsup$ of a sequence" and "taking $\liminf$ of a sequence" are both functionals, i.e. linear functions from a vector space whose elements are sequences to its underlying field of real numbers. (You can also consider it a functional in a more advanced way, by putting a topology on the vector space for which the operation is continuous.)

This captures just the additivity. To capture multiplicativity, you need to consider the space of sequences not just as a vector space, but as a ring (as before, you can put a topology on this ring, in which case it becomes natural to consider a space of sequences satisfying enough conditions to become a Banach algebra). Then $\limsup$ and $\liminf$ become ring homomorphisms from a ring whose elements are sequences to the real numbers.

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  • $\begingroup$ Aren't $\limsup/\liminf$ of a sequence different that the limit of a function? $\endgroup$ Commented Feb 1, 2020 at 16:16
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    $\begingroup$ Not when the limit exists. The reason I mention these two is that they are well-defined for all sequences, whereas $\lim$ is not. In fact, if you try to extend the definition of limit to all sequences, all possible extensions interpolate between liminf (the smallest extension) and limsup (the largest). $\endgroup$
    – pre-kidney
    Commented Feb 1, 2020 at 16:16
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    $\begingroup$ Umm, $\liminf$ and $\limsup$ aren't additive in general - nor does either one respect multiplication by negative scalars. (Plus, you would need to restrict to the space of bounded functions in order to avoid having values of $+\infty$ or $-\infty$.) $\endgroup$ Commented Feb 1, 2020 at 16:23
  • $\begingroup$ @DanielSchepler yes, which is why I was careful to avoid saying that the vector space consists of all sequences in my answer. But thanks for pointing it out explicitly. You do raise a good point though that we are not gaining much by going from lim to limsup/liminf since we will need to consider restrictions either way. $\endgroup$
    – pre-kidney
    Commented Feb 1, 2020 at 16:33

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