Question about the meaning of finite complement topology. [duplicate]

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I want to eventually show that the "finite complement topology" is in fact a topology. But first I want to make sure I understand the definition. My book says:

Let X be an infinite set. Show that $\mathscr{T}_1=\{U \subseteq X : U = \emptyset $ or $ X\setminus U $ is finite $ \}$

note: It seems very paradoxical, to me, to say $U$ is $X \setminus U$. That is like saying "$U$ is not $U$". Then I found on wikipedia "a cofinite subset of a set X is a subset A whose complement in X is a finite set. In other words, A contains all but finitely many elements of X. "

So I want to be sure that I'm understanding what this means; writing a couple specific examples:

(1) If $X = \{1,2,3...\}$, then some examples of $U$ would be
(a){2,3,4...} because the complement is {1}
(b){5,6,7...} because the complement is {1,2,3,4}
(b){1,3,5,7}$\cup${9,10,11...} because the complement is {2,4,6,8}

Or

(2) If $X = (0,10)$, then some examples of $U$ would be
(a)(0,1)$\cup$(1,10) because the complement is {1}
(b)(0,1)$\cup$(1,2)$\cup$(2,3)$\cup$(3,4)$\cup$(4,10) because the complement is {1,2,3,4}

Or

(3) If $X = \mathbb{R} ^2$ , then some examples of $U$ would be
(a) $U$ = points without integer coordinates OR any point with coordinates greater than 10, because the complement would be points with integers that are less than 10.

I know that none of those are complete list, but are those examples correct? Is it even possible to write a complete list of $U$'s for any given set?

Before I even begin this proof, I want to be sure I understand what is being asked.

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2 Answers

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As an addition: you could define a topology by what its closed sets are (instead of its open sets) and check the axioms for those.

This is a bit more natural for this topology: the closed sets are exactly all finite subsets of $X$ and $X$ itself (which is not finite, but must be closed in any topology).

Then the axioms for closed sets are quite clear: they must be closed under intersections (which is clear, as the set $X$ does nothing in intersections, and if one set in the intersection is finite, so will the intersection be) and finite unions (also clear).

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Yes, those are all correct. Note that nowhere in the definition is it said that $U$ is the same as $X \backslash U$. The definition is saying that the open sets of the topology are precisely 1) the empty set, and 2) those subsets of $X$ whose complements are finite. That's why it's called the finite complement topology. Then it becomes straightforward to verify the definition of topology using DeMorgan's laws. Your intuition for why it is a topology is as follows: finite unions of finite sets are still finite, and so are infinite intersections.

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