Consider the local base over the space of complex continuous functions over $[0,1]$ (denoted by $\mathcal{C}[0,1]$) defined for each fixed $x\in [0,1]$ and $\epsilon>0$ by
$$\mathcal{B}_{\epsilon,x} = \{f\in \mathcal{C}[0,1] : |f(x)| < \epsilon\}.$$
In an effort to better understand what is going on with the topology induced by this local base, I am interested in finding examples of a sets which is not simply equivalence classes of the zero function on $[0,1]$ or constant functions that is bounded in a topological sense (assuming such a set even exists).
My observations:
The definition of topological boundedness is really confusing me. That is, a set $S$ is bounded with respect to a topology $\tau$ if there is a $s$ such that
$$S\subseteq t\mathcal{B}_{\epsilon,x}$$
whenever $t>s$ for every open neighborhood of the origin.
Since this must be true for every single fixed $\epsilon$ and $x$, in particular as $\epsilon$ gets small and $x$ varies, the only function that would live inside all of the $\mathcal{B}$'s should be the zero function I think. Even if I were to consider only a single $\epsilon$ and $x$, the existence of a singular $t$ associated with this $\mathcal{B}$ that scales some collection of each elements $f\in \mathcal{B}$ into $S$ seems very strong if $|S|>1$.
Because each element of the local base only cares about the function value at a single point, it seems like each local base would look very different and scaling by $t$'s is unlikely to get set containment. When I consider such a set $S$ that must have this property for every neighborhood, it seems like such an incredibly strong condition that I can't fathom what they are.