I'm trying to prove the following proposition:
Let $f:X\to Y$ be a map of topological spaces. Let $F$ be a prefilter on $X$. Then
$(a)$ $f(F)$ is a prefilter on $Y$.
$(b)$ If $F$ is an ultra prefilter, then so is $f(F)$.
Note: Relevant definitions are included below.
Part $(a)$ is straightforward. But I have tried a few times to prove $(b)$ and I am stuck. I think the argument should go something like:
- Assume $F$ is an ultraprefilter.
- Suppose that $G$ is a prefilter on $Y$ which refines $f(F)$ (but is not refined by $f(F)$)
- Pass from $G$ to a prefilter $F'$ on $X$ which refines $F$ (but is not refined by $F$)
- Obtain a contradiction and arrive at the desired conclusion.
Assuming that this idea is correct, the third step (constructing $F'$) is where I am stuck.
I've tried setting $F':= \{A\subset X : f(A)\supset B\text{ for some }B\in G\}$ but this turns out not to be a pre-filter (or at least I couldn't prove that it was and I conjecture that in general it may not be).
I've also tried setting $F': \{A\subset X : f(A)\in G\}$, but I think this also fails to be a prefilter.
To sum things, I'm stuck and looking for a hint of some sort.
Thanks so much!
Definitions:
Prefilter on a topological space $X$ - A collection $F$ of subsets of $X$ such that $A_{1}, A_{2}\in F$ implies that there is $A_{3}\in F$ such that $A_{3}\subset A_{1}\cap A_{2}$. This generates the filter $\{B\subset X : A\subset B\text{ for some }A\in F\}$
Filter subbase on $X$ - A collection $I$ of subsets of $X$ which is closed under finite intersections. This generates the prefilter consisting of all finite intersections of elements of $I$.
Ultra prefilter - A prefilter which generates some ultrafilter on $X$.
Note: This is proposition 5.13 from: http://math.uga.edu/~pete/convergence.pdf