What does it mean for a sequence $x_k \in E$ such that if $x_k \to x$, then $f(x_k) \to f(x)$?

Question

This is from a homework problem, but this is not itself the homework problem. I'm trying to understand what everything means. Please be patient as I'm an on and off again student and I just keep feeling hopelessly lost and too embarrassed to talk to anyone else.

What does it mean for a sequence $x_k$ $\subseteq$ E such that if $x_k \to x$, then $f(x_k) \to f(x)$?

Edit: I mean I know what it means, that the sequence is converging onto x and so the function is converging onto x. I'm trying to see what is happening in this statement. To visualize it.

Thanks for the editing help!

Answer 1

This is an equivalent formulation of the idea of a function being continuous (under nice enough conditions; for example, functions $f:\mathbb{R}\to\mathbb{R}$ are continuous iff they satisfy the given property).

First, how to think about this statement? Suppose we have a sequence of points $(x_n)_{n\in\mathbb{N}}$ which converges to some point $x$; this means that if we fix a given tolerance $\varepsilon$, we can ensure that the distance between $x_n$ and $x$ is always smaller than $\varepsilon$ if we look at points "far enough out" in the sequence. Formally, for every tolerance $\varepsilon>0$, there exists an index $N\in\mathbb{N}$ such that for all $n\geq N$ (i.e. all points far enough out in the sequence), the distance between $x_n$ and $x$, i.e. $|x_n-x|$, is less than $\varepsilon$.

Now, from this sequence $(x_n)_{n\in\mathbb{N}}$, we can construct a new sequence $(y_n)_{n\in\mathbb{N}}$, where $y_n=f(x_n)$ for each $n\in\mathbb{N}$. This is just giving a new name to the sequence $\big(f(x_n)\big)_{n\in\mathbb{N}}$, but it is to emphasize that the objects $f(x_n)$ are just elements of a sequence $\big(f(x_n)\big)_{n\in\mathbb{N}}=(y_n)_{n\in\mathbb{N}}$. The statement that $f(x_n)\to f(x)$ is now equivalent to saying that $y_n\to f(x)$, i.e. for every tolerance $\varepsilon$ there is an $N'\in\mathbb{N}$ such that for $n\geq N'$, we have $|y_n-f(x)|<\varepsilon$.

The implication $x_n\to x\implies f(x_n)\to f(x)$ means that, given any sequence $(x_n)_{n\in\mathbb{N}}$ which converges to some point $x$, if we look at the sequence $(y_n)_{n\in\mathbb{N}}=\big(f(x_n)\big)_{n\in\mathbb{N}}$, we have that $y_n\to f(x)$. That is, the sequence formed by applying $f$ to each element $x_n$ of the original sequence converges to $f(x)$, the value we get when we apply $f$ to the original limit $x$.

Again, this is just a reformulation of the idea of continuity. It's a good exercise to show that this is equivalent to the usual $\varepsilon-\delta$ formulation of continuity. It's also good to play around with a few examples to see why this property characterizes continuous functions. For example, if we take $f:[0,1]\to\mathbb{R}$ where $f(x)=0$ if $x\neq0$ and $f(x)=1$ if $x=0$, this function would clearly not be continuous at $0$ from the $\varepsilon-\delta$ characterization. Instead, if we look at the characterization of continuity via sequences, to prove that $f$ is not continuous at $0$, we'd need to exhibit a sequence $(x_n)_{n\in\mathbb{N}}$ converging to $0$ such that $f(x_n)\not\to f(0)$. It we take $x_n=1/n$, for example, then $x_n\to0$ and $f(1/n)=0$ for each $n\in\mathbb{N}$, but $f(0)=1$, so $f(x_n)\not\to f(0)$. Hopefully this example helps you to visualize what is going on in the definition.