How does one extend the limit concept to extended real numbers?

Question

Consider a function $f: \mathbb{R}\to [-\infty, + \infty]$. Is it possible to define

$$\lim_{x \to a} f(x)$$

where $a \in \mathbb{R}$ without appealing to a metric on $[-\infty, + \infty]$?

If yes, how?

Answer 1

Eric Yau's comment goes straight to the point: the extended real line $\overline{\mathbb{R}} = [-\infty, +\infty]$ is a topological space, and it is common knowledge how to extend the definition of $\lim_{x\to a}f(x) = b$ for functions $f \colon X \to Y$ between metric spaces to the case where $Y$ (and possibly also $X$) is a topological space on which a metric may not have been given.

An answer to the question can be assembled from the Wikipedia page Limit of a function [which incidentally needs some tidying up]:

Functions on topological spaces

Suppose $X,Y$ are topological spaces with $Y$ a Hausdorff space. Let $p$ be a limit point of $\Omega \subseteq X$, and $L \in Y$. For a function $f \colon \Omega \to Y$, it is said that the limit of $f$ as $x$ approaches $p$ is $L$ (i.e., $f(x) \to L$ as $x \to p$) and written $$ \lim_{x \to p}f(x) = L $$ if the following property holds:

• For every open neighborhood $V$ of $L$, there exists an open neighborhood $U$ of $p$ such that $f(U \cap \Omega − {p}) \subseteq V$.

This last part of the definition can also be phrased "there exists an open punctured neighbourhood $U$ of $p$ such that $f(U \cap \Omega) \subseteq V$".

Limits involving infinity

[...]

• a neighborhood of $-\infty$ is defined to contain an interval $[-\infty,c)$ for some $c\in\mathbb{R}$,
• a neighborhood of $\infty$ is defined to contain an interval $(c,\infty]$ where $c\in\mathbb{R}$, and
• a neighborhood of $a\in\mathbb{R}$ is defined in the normal way [for the] metric space $\mathbb{R}$.

[...] $\overline{\mathbb{R}}$ is a topological space and any function of the form $f\colon X\to Y$ with $X,Y\subseteq\overline{\mathbb{R}}$ is subject to the topological definition of a limit. Note that with this topological definition, it is easy to define infinite limits at finite points, which have not been defined above in the metric sense.

I said that this is "common knowledge", but it seems remarkably hard to find a clear statement of it in print! (One could always print out the Wikipedia page, I suppose!) :)

As indicated in my comment, I thought that the definition of the limit of a function between general topological spaces might be buried somewhere in a discussion of filter convergence (or net convergence); but I couldn't find it there, either (and I looked in quite a number of books). I hope that someone can supply the lack of an authoritative reference.

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A clear general definition of the meaning of $\lim_{x\to x_0} f(x) = y$ can be given by condensing extracts from pages 48--62 of Horst Schubert, Topology (Macdonald 1968):

A non-empty system $\mathscr{B}$ of subsets of the set $X$ is called a filter basis (on $X$) if the following conditions are satisfied:

FB 1 The intersection of two sets from $\mathscr{B}$ contains a set from $\mathscr{B}$.

FB 2 The empty set does not belong to $\mathscr{B}$.

A non-empty system $\mathscr{F}$ of subsets of the set $X$ is called a filter (on $X$) if the following conditions are satisfied:

F 1 Every superset of a set from $\mathscr{F}$ belongs to $\mathscr{F}$.

F 2 The intersection of two sets from $\mathscr{F}$ belongs to $\mathscr{F}$.

F 3 The empty set does not belong to $\mathscr{F}$.

The neighbourhoods of a point $x$ form a filter, the neighbourhood filter of $x$. Every filter basis $\mathscr{B}$ on $X$ determines a filter $\mathscr{F}$ on $X$.

Let $\mathscr{F}$ and $\mathscr{G}$ be two filters on the set $X$. If every set of $\mathscr{F}$ belongs to $\mathscr{G}$, then $\mathscr{F}$ is said to be coarser than $\mathscr{G}$ and $\mathscr{G}$ finer than $\mathscr{F}$.

Let $X$ be a topological space. A filter $\mathscr{F}$ on $X$ is said to be convergent to a point $x$ if $\mathscr{F}$ is finer than the neighbourhood filter of $x$. If $\mathscr{F}$ converges to $x$ then $x$ is called a limit of $\mathscr{F}$ and one writes $\mathscr{F} \to x$ or $x \in \lim \mathscr{F}$. One says that $x = \lim \mathscr{F}$ if $x$ is the only limit of $\mathscr{F}$. A filter basis $\mathscr{B}$ is said to be convergent to $x$ if the filter with the basis $\mathscr{B}$ converges to $x$.

Let $X$ be a topological space. The following assertions are equivalent:

H Every two distinct points of $X$ have distinct neighbourhoods.

H$'$ Every point of $X$ is the intersection of its closed neighbourhoods.

H$''$ The diagonal of $X \times X$ is closed.

H$'''$ There exists no filter on $X$ that converges to two different points.

A topological space $X$ is said to be Hausdorff or separated if it satisfies one (and hence all) of these conditions.

Let $f \colon X \to X'$ be a mapping [function] and $\mathscr{F}$ a filter on $X$. The filter on $X'$ that has images of the sets from $\mathscr{F}$ as a basis is called the image of $\mathscr{F}$ relative to $f$ and will be denoted by $f(\mathscr{F})$.

If $\mathscr{B}'$ is a filter basis on $X'$, then the inverse image of sets from $\mathscr{B}'$ relative to $f$ satisfy condition FB 1. Condition FB 2 is satisfied if and only if no set from $\mathscr{B}'$ has an empty inverse image.

If $\mathscr{F}'$ is a filter on $X'$ and the inverse images of sets from $\mathscr{F}'$ are non-empty, then these images form, in general, only a filter basis.

Let $f \colon X \to X'$ be a mapping and $\mathscr{F}'$ a filter on $X'$. If the inverse images of sets from $\mathscr{F}'$ form a filter basis on $X$, then the filter with this basis is called the inverse image of $\mathscr{F}'$ relative to $f$ and will be denoted by $f^{-1}(\mathscr{F}')$.

Let $\mathscr{B}'$ be a filter basis on $X'$. The filter $\mathscr{F}'$ with basis $\mathscr{B}'$ has an inverse image relative to the mapping $f \colon X \to X'$ if and only if no set from $\mathscr{B}'$ has an empty inverse image.

The following important special case is covered by these considerations. Let $A$ be a subset of $X$. Relative to the inclusion of $A$ in $X$ every filter $\mathscr{G}$ on $A$ has an image on $X$ which in this case is called the extension of $\mathscr{G}$ to $X$. If $\mathscr{F}$ is a filter on $X$, then the inverse image of $\mathscr{F}$, relative to the inclusion of $A$ in $X$, need not exist. It exists if and only if $A$ meets every set from $\mathscr{F}$. This inverse image is also called the trace of $\mathscr{F}$ on $A$.

We introduce some more terminology and assume for this purpose that $Y$ is Hausdorff. The continuity of the mapping $f \colon X \to Y$ at the point $x_0$ is equivalent to $$ \lim_{x\to x_0} f(x) = f(x_0), $$ which asserts that the image of the neighbourhood filter of $x_0$ converges to $f(x_0)$. If the neighbourhood filter of $x_0$ induces a filter $\mathscr{F}$ on a subset $A$ of $X$ [it is clear that by this form of words he means that $\mathscr{F}$ has a trace on $A$], then we write $$ \lim_{\substack{x\to x_0\\x\in A}} f(x) = y, $$ when the image of $\mathscr{F}$ relative to the mapping $f|_A$ converges to $y$. If in particular $A = X \setminus \{x_0\}$, then we will also write $$ \lim_{\substack{x\to x_0\\x\ne x_0}} f(x) = y. $$ [It is clear that this notation is defined if and only if $x_0$ is not an isolated point of $X$; and then $\mathscr{F}$ is the collection of all punctured neighbourhoods of $x_0$ in $X$.]