There must be an error in the following argument since Fin is not many-one reducible to Inf, I can't seem to find it. Here it is informally (I hope it's straightforward and not confusing):
Take any infinite r.e. set like $K$ and consider the following:
We will enumerate members of $W_{e}$ and $K$ (simultaneously). We start with $A^0=\emptyset$ and at each stage $i$ given $A^i$ if we've found a new element in $K$ then we add it to $A^{i+1}$. If at this stage we also find a new element in $W_{e}$ then we let $A^{i+1}=\emptyset$ otherwise let $A^{i+1}=A^i\cup$ whatever new element we found in $K$ (if one was found). If we found a new element of $W_{e}$ at stage $i$, then we begin enumerating $K$ from scratch at the next stage, otherwise we keep enumerating $K$ where we left off. Regardless of what happens at stage $i$ we keep enumerating $W_{e}$ where we left off in the next stage.
At the end let $W_{g(e)}=\bigcup_{i<\omega}A^i$
What will then happen is:
1) if $W_{e}$ is finite then at some point no new elements of $W_{e}$ will be found and so we will be able to place all elements of $K$ into $W_{g(e)}$ without $W_{e}$ hindering us. So in the end we will have $W_{g(e)}=K$, which means $W_{g(e)}$ will be infinite.
2) if $W_{e}$ is infinite, then we will always be finding new elements of $W_{e}$ and we will be put in a loop and never be able keep anything inside of $W_{g(e)}$. So in the end we will have $W_{g(e)}=\emptyset$, which means $W_{g(e)}$ will be finite.
Any help is appreciated, thanks.
When you say that you are trying to make $W_{g(e)}$, you are giving a computable enumeration of a set that should be c.e. in the end. A c.e. enumeration can only put elements into a set, but can not take it out. In your construction, "Each time a new element is found in $W_e$ we clear out everything we have in $W_{g(e)}$" This involves removing elements from your enumeration.
Such a construction of a set which merely has an computable approximation where elements come in and out are not necessarily c.e. sets. There is a general notion of a $n$-c.e. set. These are sets that possess a computable approximation where for each element $x$, it can come in and out of the approximation at most $n$ times. In this sense, the 1-c.e. sets are the ordinary c.e. sets. It is very easy to show that there is a 2-c.e. set that is not c.e.