Is ∞ a limit point of $\mathbb R$ ? If not, how to understand Rudin's definition at the beginning of chapter $4$ (PMA)?

Question

I am starting reading the $4^{th}$ chapter of PMA from Walter Rudin.

The chapter is about continuity and it defines $$\lim_{x\to a} f(x)$$ (for a function mapping a metric space $E$ into a metric space $F$) saying $a$ is a limit point of $E$.

I don't understand: for a function whose domain is $\mathbb R$, does it mean that for $\lim_{x\to \infty} f(x)$ we have ∞ being a limit point of $\mathbb R$ ?

According to this post, $\mathbb R '=\mathbb R$, and ∞ is not a limit point of $\mathbb R$.

PS: Here is the (beggining of the) definition:

Answer 1

Let's start with a technical point about topology. If we declare that our universe is the set $X$, then we don't care about any sets that might contain $X$ in another universe. This means that when we consider the closure of $X$, we are automatically constrained to $X$ itself. There is nothing else besides. Formally, when we turn $X$ into a topological space $(X, \tau)$, then it's necessarily true that $X$ is closed with respect to $\tau$.

If our universe is $\mathbb{R}$, then $\mathbb{R}$ contains all of its limit points. The point $+\infty$ simply doesn't exist. To make the point a bit firmer, if our universe was instead the rationals $\mathbb{Q}$, then the point $\pi$ simply doesn't exist. On the other hand, if we consider $\mathbb{Q}$ as a subset of $\mathbb{R}$, then we can show that $\pi \in \mathbb{R} \setminus \mathbb{Q}$ is indeed a limit point of $\mathbb{Q}$ (e.g., take an expanding decimal representation).

This is all an intuitive picture for the following fact: a set $A \subset X$ is always closed relative to the subspace topology $\tau_A = \{A \cap U : U \in \tau\}$. This means that the closed sets can change when we change our universe. $\mathbb{Q}$ isn't closed when the universe is $\mathbb{R}$, but it is closed when the universe is $\mathbb{Q}$.

Now to the next point. When we study functions $f : \mathbb{R} \to \mathbb{R}$, we are sometimes interested in the long run behavior of $f(x)$ as $x$ becomes very large (or very small). There is no point in $\mathbb{R}$ called "the long run", which means we have to extend a notion of limits without one. When we write $$ \lim_{x \to +\infty} f(x) = L \tag{$\star$} $$ we aren't asking about $|+\infty-x|<\delta$, because that statement has no meaning. To capture the "infinite limit" idea, we have to use a different approach. Instead, we say that $x \to +\infty$ means that eventually, $x$ is greater than any number $M$. So we impose the following interpretation of ($\star$): for every $\varepsilon > 0$ there exists $M \in \mathbb{R}$ such that if $M < x$, then $|f(x)-L|<\varepsilon$. This captures our meaning of "long run behavior" without having to admit $+\infty$ into our space.

Finally, we might ask if there is a space $\mathbb{R} \cup \{+\infty\}$ with a topological structure that describes both our conventional limits $x \to x_0$ and our extended limits $x \to +\infty$. The answer is yes.

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Answer 2

Rudin introduces the extended real number system $\mathbb R \cup \{-\infty, \infty \}$ in Definition 1.23 on p.11. Let us denote it by $\overline{\mathbb R}$.

In the section entitled "Infinite limits and limits at infinity" (p.97 ff) he explains how to work in the extended real number system:

... we shall now enlarge the scope of Definition 4.1 by reformulating it in terms of neighborhoods.

This is made explicit in Definition 4.33 and answers your question what $\lim_{x \to \infty} f(x)$ means.

It does not mean that $\infty$ is a limit point in the metric space $\mathbb R$ (this is impossible because $\infty \notin \mathbb R$). It has a precise meaning in terms of neigborhoods of $\infty$ in $\overline{\mathbb R}$. Actually $\infty$ can be regarded as a limit point in $\overline{\mathbb R}$, although not in the sense of Definition 2.18 (because Rudin does not introduce a metric on $\overline{\mathbb R}$).