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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

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2 Answers 2

Let $\mathscr{F}$ be the ultrafilter generated by $F$. Let $\mathscr{G}$ be the filter generated by $f[F]$, and let $A$ be any subset of $Y$. Let $B=f^{-1}[A]$. Since $\mathscr{F}$ is an ultrafilter, either $B\in\mathscr{F}$ or $X\setminus B\in\mathscr{F}$. If $B\in\mathscr{F}$, there is some $H\in F$ such that $H\subseteq B$, in which case $f[H]\subseteq f[B]\subseteq A$, and $A\in\mathscr{G}$.

Similarly, if $X\setminus B\in\mathscr{F}$, there is some there is some $H\in F$ such that $H\subseteq X\setminus B$, in which case $f[H]\subseteq f[X\setminus B]\subseteq Y\setminus A$, and $Y\setminus A\in\mathscr{G}$. Thus, for any $A\subseteq Y$, $\mathscr{G}$ contains either $A$ or $Y\setminus A$ and is therefore an ultrafilter.

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I recommend that you show the following, instead, as I think you'll have an easier time of it (and the results are useful, themselves): (1) If $F$ is a prefilter and $\mathcal{F}$ the filter generated by $F$, then $f(F)$ and $f(\mathcal{F})$ are prefilters that generate the same filter. (More generally, if $F,F'$ are prefilters that generate the same filter, then $f(F),f(F')$ have the same property.) (2) If $\mathcal{F}$ is an ultrafilter, then $f(\mathcal{F})$ is an ultra prefilter.

If you'd like any prompts for those two results, just let me know.

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