# For a function $f: X to Y$, if $Y-V$ is finite, when is $X - f^{-1}(V)$ finite?

Mathematics Asked by trujello on August 17, 2020

I apologize if this is a silly question but I just do not know enough set theory (i.e., sizes) to understand if it’s even silly.

My question is

Let $$f: X to Y$$ be a function of sets. Suppose $$V subset Y$$, and that $$Y – V$$ is finite (countable). Is $$X – f^{-1}(V)$$ finite (countable)? What conditions do we need to place on $$f$$ to guarantee it will be finite (countable)?

So I kind of have two questions. But I will take either.

Why I care:

I’m asking this because I want to know the following. Recall that for a set $$X$$, we can endow $$X$$ with the finite complement topology where a set $$U subset X$$ is open if $$X – U$$ is finite. Denote this topology on a set $$X$$ as $$FC_X$$.

Now suppose $$f: X to Y$$ is a function. As $$X$$ and $$Y$$ are sets, we may ask:

Does the function $$f: X to Y$$ extend to a continuous function $$f: (X, FC_X) to (Y, FC_Y)$$?

For such a function to be continuous, we need that $$f^{-1}(U)$$ is open if $$U$$ is open. In this case, that means that if $$Y – U$$ is finite then $$X – f^{-1}(U)$$ must be finite. But I do not know when that last statement holds, and hence it’s my question above. My guess is that if $$f$$ is injective, then it will be true, but I don’t really know if that’s true. I lack the set theory knowledge to really attack such a problem.

Finally, if the above answer is true, it tells me that I’ve got a functor $$F: textbf{Set} to textbf{Top}$$ where $$X$$ is sent to the finite complement topology. I’m also interested in the countable case, since that would give me another different functor. But that’s besides the question.

As pointed out by @Cronus in the comments, it is sufficient that $$f$$ has the property that the preimage of every point is a finite set. I believe it is also a necessary condition.

Let's just consider surjective functions, just for convenience (it really doesn't change anything since any function $$f:X to Y$$ can be 'made surjective' by replacing its codomain with its image.)

To see that the condition is sufficient, note that if some $$f: X to Y$$ has the property, then for finite $$Y setminus V$$, $$X setminus f^{-1}(V) = f^{-1}(Y setminus V) = bigcup_{y in Y setminus V} f^{-1}(y).$$ In other words, $$X setminus f^{-1}(V)$$ is the union of a finite number of finite sets.

I believe this condition is also necessary, and here is a proof that seems to be correct. Assume that $$f$$ doesn't have the property, and thus there exists some $$y$$ with $$f^{-1}(y)$$ not finite. Taking $$V = Y setminus { y }$$, one then concludes that $$X setminus f^{-1}(V) = f^{-1}(y),$$ which is not finite. Importantly, we have that $$f^{-1}(V)$$ does not intersect $$f^{-1}(y)$$ since no $$x in X$$ can be mapped both into $$V$$ and into $${ y }$$. Thus by contradiction we have that $$f$$ must have the finite-to-one property.

Correct answer by Jacob Maibach on August 17, 2020

Let $$X$$ be an infinite set, $$Y$$ a finite set and $$V = emptyset$$. Then for any function $$f : X to Y$$ we have that $$Ysetminus V = Y$$ is finite but $$X setminus f^{-1}(V) = X$$ is infinite.

We get a similar conclusion for your other question if $$X$$ is an uncountable set and $$Y$$ is a countable set.

Answered by mechanodroid on August 17, 2020

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