Functions representable as a sum of a continuous function and a periodic function, and the axiom of choice.

36 Views Asked by At

Let $A$ be the set of continuous functions from the reals to the reals, let $P$ be the set of periodic functions from the reals to the reals, and $F$ be the set of all functions from the reals to the reals. Let $S$ the set of functions which are expressible as the sum of an element from $A$ and an element from $P$. Note that $A$ has the same cardinality as $\mathbb{R}$, but $P$ has the same cardinality as $2^{|\mathbb{R}|}$ which is the cardinality of $F$. So there is no cardinality based argument that there are elements in $F$ which are not in $S$.

However, it seems like there should be elements in $F$ which are not in $S$, and my guess is that there should be some construction possibly using a Hamel basis.

In that context I have three questions:

  1. Can we show there is an element of $F/S$?
  2. Can we prove 1 without assuming the Axiom of Choice?
  3. Do the answers for 1 and 2 change if in $A$ we allow instead of just continuous functions, functions which are continuous except at countably many points?