Context: I was solving an integral which had a typo that made it impossible to solve and when I wrote it on wolfram it gave me "no result found in terms of standard mathematical functions" this made me wonder how many functions that we need to define so that we can solve every integral or express anything with.
Let $S$ be the set that contains almost everywhere continuous functions on $\mathbb{R}$ such that if $f,g \in S$ then $f+g, \, f\times g, \ f^g, \ f(g) \in S$ , $af, \ f^a \in S \ \forall a \in \mathbb{R}$ provided they are defined for all $x$ .
My question is How many elements should $S$ have such that any almost everywhere continuous real function could be expressed as a finite combinations of elements of $S$.
By combinations here I mean that $f+g, \, f\times g, \ f^g, \ f(g), af, \ f^a \ \forall a \in \mathbb{R}$
Of course $S$ should be an infinite set but could it be countable set and still could express any continuous function with its elements?
I think it is impossible for such set to exist but I don't know how to prove or disprove it.
Best Answer
The set of finite combinations of $S$ has cardinality $\mathfrak{c}$ (the continuum). But the set of almost-everywhere continuous functions has cardinality $\mathfrak{c^c}$, which is greater. In fact, let $E$ be the Cantor set, and consider real-valued functions which are $0$ on the complement of $E$ and arbitrary on $\mathbb E$. These are continuous on the complement of $E$ (and thus almost everywhere), and their cardinality is $\mathfrak{c^c}$. Thus not all of these can be expressed as finite combinations of $S$.