How to visualize the open ball in $C[0,1]$?

Question

Consider the metric space $C[0,1]$ with 'sup' metric: $$d_{\infty}(f,g)=\text{sup}_x\;\vert f(x)-g(x)\vert $$

How to visualize the open ball $B(f,r)$ where $f$ is the identity function and $r>0$?

Here, $B(f,r)=\{g \in C[0,1]: \text{sup}_x\;\vert x-g(x)\vert<r\}. $

Is the figure look like the tube around the line $y=x$ ?

Added:

By DanielWainfleet answer, the figure look like this: (An outline with $r=1$)

Where the middle line is the identity function and and the upper line is the graph of $1+x$ and the bottom line is $-1+x$ .

A function $g \in B(f,1)$ means $g$ lies entirely within this strip!

Answer 1

Picture the graph of a continuous $f:0,1]\to \Bbb R$ and the graphs of $g(x)=r+f(x)$ and of $h(x)=-r+f(x).$ Continuous functions whose graphs lie entirely below $g$ and above $h$ are the members of $B(f,r) .$ In your Q a member $g$ of $B(id_{[0,1]},r)$ satisfies $-r+x<g(x)<r+x$ for every $x.$

BTW, this is sometimes called the metric of uniform convergence because $\lim_{n\to \infty}d_{\infty}(f_n,f)=0$ iff the sequence $(f_n)_n$ of functions converges uniformly to $f.$