Calrifying a theorem on Cauchy sequences

Question

What does it mean for a sequence to be in the space of bounded continuous functions and (the supremum norm of all functions?-- not sure what the thing in red means exactly). Also, by definition doesn't every Cauchy sequence converge?

Answer 1

A sequence $(a_n)_{n\in\mathbb{N}}$ lies in the space of bounded continous functions if every $a_n$ is a bounded continous function.

[similar to: $(a_n)_{n\in\mathbb{N}}$ is a sequence in the real numbers if every $a_n \in \mathbb{R}$]

It means every $a_n$ is a continous function $a_n : A \to \mathbb{R}$ and is bounded. But what does bounded mean in this context? To a given norm $\|\cdot\|$ on $C(A)$ a function $a_n \in C(A)$ is called bounded if $\|a_n\| < \infty$.

You write $(C(A),\|\cdot\|_\infty)$ so that everybody knows you are in the normed vector space of continous functions $A\to\mathbb{R}$ and use $\|\cdot\|_\infty$ as your norm. And therefore you see $C(A)$ as a metric space with the metric induced by $\|\cdot\|_\infty$ [which is important for you cauchy sequences].

While in finitly dimensional vector spaces like $\mathbb{R}^n$ all norms are equivalent and therefore give you the same open sets, convergent series, etc. it can make a difference which norm you use on an infinitely dimensional vector space like $C(A)$. Therefore it‘s important to emphasize which norm you use.

Cauchy sequences don’t always converge! It‘s important to emphasize in which metric space your sequence lies. You can construct examples where it doesn’t work. For example a cauchy sequence in $\mathbb{Q}$ which would converge to $\sqrt{2}$ in $\mathbb{R}$ but diverges in $\mathbb{Q}$ since $\sqrt{2} \not\in \mathbb{Q}$. A metric space in which every cauchy sequence converges is called complete. Examples for complete metric spaces are: $(\mathbb{R}, |\cdot |),\ (\mathbb{R}^n, \|\cdot\|_2), (\mathbb{R}^n,\| \cdot \|_\max)$ and as your lemma shows $(C(A),\|\cdot\|_\infty)$.