Topology/Lesson 4

Introduction to Limits

This lesson will introduce the notion of a limit.

Definition (sequence)

A sequence on a topological space $X$ is a function $f:\mathbb{N}\to X$. Alternatively, it is a list $x_1,x_2,x_3,\dots$ where $x_i\in X$ for all $i$. The sequence is often denoted as $\{x_n\}$

Definition (limit of a sequence)

The point $x\in X$ is a limit of the sequence $\{x_n\}$ if for every neighborhood $N\ni x$, there is $K\in \mathbb{N}$ such that $x_n\in N$ for all $n\ge K$. In this case, we write $x_n\to x$ and say that $\{x_n\}$ converges to $x$.

Note that a sequence might have multiple limits. For example, in any space with the indiscrete topology, every sequence converges to every point of the space!

Example

Let $X=\mathbb{Z}$ be the set of integers with the topology where $U\subset X$ is open if $|U^c|<\mathrm{\infty }$ (called the finite complement topology). Let $x_n=n$. Then we see that $x_n\to x$ for any $x\in X$. Indeed, note that given any neighborhood $N\ni x$, $N$ contains all but finitely many points of $X$. Let $K$ be the maximum of all of the numbers not contained in $N$. Then for all $n>K$, we see that $x_n=n\in N$, hence $x_n\to x$.

Theorem

If a space $X$ is Hausdorff and the sequence $\{x_n\}$ in $X$ has a limit, then that limit is unique.

The proof of this theorem is left as an exercise to the student. The hint is to assume that there are two distinct limits and show that this leads to a contradiction.

Theorem

If $f:X\to Y$ is a continuous function and $x_n\to x$ in $X$ then $f(x_n)\to f(x)$ in Y.

Proof

Let $N\ni f(x)$ be an open neighborhood of $f(x)$. Since $f$ is continuous, $N^{\prime }=f^{-1}(N)$ is open in $X$ and, by definition, contains $x$. Therefore, there is $K\in \mathbb{N}$ such that $x_n\in N^{\prime }$ for all $n>K$. Therefore, $f(x_n)\in f(N^{\prime })=N$. Thus, $f(x_n)\to f(x)$.

The converse of this theorem is, in general, false. However, it is true for metric spaces. (In fact, it holds for any first-countable space.)