# Show “countable complement topology” is a topology

I'm teaching my self topology with the aid of a book. I'm trying to prove the following is a topology:

Let X be an infinite set. Show that $\mathscr{T}_2=\{U \subseteq X : U = \emptyset$ or $X\setminus U$ is countable$\}$ is a topology.

My book calls this set the "countable complement topology"

Please let me know if my proof is valid.

corollary [A]: The union of countable sets are countable.

proof: I'm going to take as a given that that the union of finite set are finite, and just prove the case of countable infinite sets. A set is countable infinite if there is a bijection between the set and the natural numbers. Let $A$ and $B$ be a countable infinite sets. Now let $f$ be the bijection between $A$ and $\mathbb{N}$ and $g$ be $B$'s bijection.
$f:\mathbb{N} \rightarrow A$
$g: \mathbb{N} \rightarrow B$

There can always exist a bijection $h$ that maps $A \cup B$ to $\mathbb{N}$. To account for non disjoint sets, let $B' = B \setminus A$; (note: $A \cup B = A \cup B'$)
$h: \mathbb{N} \rightarrow A \cup B'$

$h(x) = \left\{ \begin{array}{ll} f(\frac{1}{2}x); & \quad x \in 2\mathbb{N} \\ g(\frac{x+1}{2}); & \quad x \in 2\mathbb{N}+1 \end{array} \right.$
$h^{-1}(x)$ would map $A$ to the even numbers, and $B'$ to the odd numbers. Therefore, The union of countable sets are countable. (QED)

Now to verify the definition.

(i) X and Ø are elements of $\mathscr{T}$.

Ø is given in the definition of $\mathscr{T}_2$. Also, the empty set is considered finite, with cardinality zero. Thus $X$ is in $\mathscr{T}_1$ because its complement is finite, and finite numbers are countable.

(ii) $\mathscr{T}$ is closed under finite intersections.

Let:$V_k \subseteq X$ be any countable set. (ie: $V_1, V_2...$ are all countable sets)
So Then, $U_k = X \setminus V_k$, would be the subsets of $\mathscr{T}_2$.

$U_k \cap U_j = X \setminus (V_k \cup V_j)$, due to De Morgan's law.

Thus, $U_k \cap U_j \in \mathscr{T}_2$, since the Union of two countable sets is also countable[A]. And This reasoning can be extended to any finite amount of intersections.

(iii) $\mathscr{T}$ is closed under arbitrary unions.

Let $A=\bigcup_{i \in I} U_i$ be an arbitrary union of sets whose complements are are countable.Then by De Morgan's law, $A=X \setminus (\bigcap_{i \in I}V_i)$, for some countable sets $V_i$. By the definition of intersection, $\bigcap_{i \in I}V_i \subseteq V_i$ for all $i \in I$. Since $V_i$ is countable so is any of its subsets. Hence the complement of $A$ is countable.

• Your corollary is slightly problematic because maybe $A\cap B\neq \varnothing$, but other arguments are fine. – cjackal Jun 24 '16 at 5:54
• I see what you mean, because then the function would not be 1-to-1. I guess i would have to add an extra condition. Hopefully I can think of a simple way to account for that. – Michael Maliszesky Jun 24 '16 at 14:56
• Note that the countable union of countable sets is also countable, so the co-countable topology has the unusual property that a countable intersection of countably many open sets is also open (where we can normally go no further than finite intersections). – Henno Brandsma Jun 24 '16 at 17:01
• I just added $B'$ for the case of non-disjoint sets – Michael Maliszesky Jun 24 '16 at 18:18
• What book are you using? I can recommend several excellent and inexpensive ones. – Mathemagician1234 Jun 24 '16 at 18:30

The proofs are almost OK. You do need to handle the special case of the empty set in your proofs.

So for (ii): suppose $U_1,\ldots,U_n$ are in $\mathscr{T}$, where $n \in \mathbb{N}$.

If one of the $U_i$ happens to be equal to $\emptyset$, then $\cap_{i=1}^n U_i = \emptyset$ as well so $\cap_{i=1}^n U_i \in \mathscr{T}$ as required.

In the other case, all $U_i$ obey that $X \setminus U_i$ is at most countable, so then de Morgan says that $X \setminus (\cap_{i=1}^n U_i) = \cup_{i=1}^n (X \setminus U_i)$, which is a finite union of (at most) countable sets so (at most) countable. Hence $\cap_{i=1}^n U_i \in \mathscr{T}$, as required.

As an aside: Note that in this case, because we know that even a countable union of (at most) countable sets is (at most) countable, we can even show that the countable intersection of open sets is open. Also the intersection of (countably many) non-empty open sets will be non-empty, in case $X$ is uncountable.

As to (iii): if $U_i, i \in I$ are in $\mathscr{T}$, then we can assume all $U_i \neq \emptyset$, because they do not contribute to the union, and we can assume we have at least one $U_{i_0}$ that is non-empty (or the union would be empty and so in $\mathscr{T}$ anyway).

Now, $X \setminus ( \cup_{i \in I} U_i ) = \cap_{i \in I} (X \setminus U_i) \subseteq X \setminus U_{i_0}$, and the last set is at most countable, so $\cup_{i \in I} U_i \in \mathscr{T}$, being co-countable as well.

I think that (as this is a topology course) claiming that a finite (or countable) union of countable sets is countable, is quite justified without giving a proof for that: such basic set theory should be a given before a topology course, in my opinion.

• A similar can also be found here. – Viktor Glombik Mar 11 at 0:41