# Open and closed balls in $C$

Let $X$ be a non empty set and let $C[a,b]$ denote the set of all real or complex valued continuous functions on $X$ with a metric induced by the supremum norm.

How to find open and closed balls in $C[a,b]$? Can we see them geometrically? For example what is an open ball $B(x_0;1)$ i.e. ball centered at $x_0$ with radius $1$ in $C[a,b]$. I can visualize them in $\mathbb R^n$ but when it comes to functional spaces I have no clue how to identify them?

Thanks for helping me.

#### Solutions Collecting From Web of "Open and closed balls in $C$"

Yes, you should think of it just like you think of any other metric space. Every norm $\|\cdot\|$ induces a metric $d(x,y) := \|x-y\|$.

$$B(x_0, 1) = \{ f: X \to \mathbb R \Big \vert \|f – x_0\|_\infty < 1 \}$$
In the $\sup$-norm, these $f$ are all functions that are never further away from $x_0$ at any given $x$ in $\mathbb R$. This is what it looks like:
So I’m new here and sorry if I have this wrong, but apparently I’m not meant to respond to other answers; yet I also need 50 reputation to add a comment, so I can’t do that? Anyway, I believe the above answers are incorrect. Just take the center of an open ball to be the zero function (radius 1) and $f$ to be $2/\pi \arctan(x)$. Then $f(x)$ is always within a distance of 1 from the center (strictly) in R, but $f$ is a distance of 1 from the zero function (exactually 1, so not in the open ball) in the function space, because the supremum of $\{f(x)-0 | x \in R\}$ is indeed 1. In fact f is on the boundry in the function space despite its values never touching 1 or -1 in R.