# topology of sets with countable complements

I am trying to write a proof for q 2 on page 94 in third chapter of Simmon's "Topology and Modern analysis"

Let $$X$$ be a non empty set. Consider a class of subsets including $$\emptyset$$ set and all subsets which have a countable complement. Is this a topology on $$X$$.

My attempt: Let the mentioned class be denoted by $$\mathscr{C}$$

Proof that arbitrary unions are included in $$\mathscr{C}$$:

Let $$A_i$$ be sets of any countable non empty subclass $$\mathscr{A}$$ of $$\mathscr{C}$$, then the complements $$A_i'$$ are countable.

Consider union of sets in $$\mathscr{A}$$ given by $$\bigcup\limits_{A_i \in \mathscr{A}}A_i$$ Its complement is given by $$\bigcap\limits_{A_i \in \mathscr{A}}A_i'$$.

Since $$\#(\bigcap\limits_{A_i \in \mathscr{A}}A_i') \le \#(A_i')$$ for all $$i$$, where $$\#()$$ is cardinality,

$$\bigcap\limits_{A_i \in \mathscr{A}}A_i'$$ should be countable

Thus, $$\bigcup\limits_{A_i \in \mathscr{A}}A_i$$ should be in $$\mathscr{C}$$ as its complement is countable.

Proof that finite intersections are included in $$\mathscr{C}$$:

Similarly, let $$\mathscr{B}$$ be a finite suclass of $$\mathscr{C}$$. Let sets in $$\mathscr{B}$$ be denoted by $$B_i$$. Then intersection of sets in $$\mathscr{B}$$ is given by $$\bigcap\limits_{B_i \in \mathscr{B}}B_i$$ and its complement by $$\bigcup\limits_{B_i \in \mathscr{B}}B_i'$$. Since this is a finite union of countable sets, it must be countable.

$$X$$ itself is in $$C$$ as $$\emptyset$$ is countable.

So we can conclude that $$\mathscr{C}$$ forms a topology.

Proof that finite union of countable sets is countable:

Let $$Y_i$$ for $$1 \le i \le n$$ be countable sets. Let $$y_{i,j}$$ represent $$j^{th}$$ element of $$i^{th}$$ class. Then elements of $$\bigcup\limits_{1\le i \le n}Y_i$$ maps into set of integers using $$f(y_{i,j})=n(j-1)+i$$. Since at least one of the sets is countable and not finite set of integers maps into elements of this class.

Since $$\bigcup\limits_{1\le i \le n}Y_i$$ maps into set of integers and vice versa, they are numerically equivalent using Schroeder Bernstein's theorem.

• minor quibble: if $\mathcal{C}$ is empty its union is $\emptyset$ so in the class too. Otherwise we have some $C_i$ in the family and indeed $(\bigcup \mathcal{C})^\complement \subseteq C_i^\complement$ hence at most countable. – Henno Brandsma May 14 '19 at 6:11
• Thanks for pointing out the mistake – Curious May 14 '19 at 6:21
• Not really a mistake just an extra case. – Henno Brandsma May 14 '19 at 6:23
• This is generally called the co-countable topology on $X$. – DanielWainfleet May 14 '19 at 9:14

Your mistake is that the Schröder-Bernstein theorem can't apply in this case since $$f$$ is not injective.
Hint: $$|\mathbb{N}| = |\mathbb{N}^2|$$, and use induction.
• I did not understand why it will not be injective. As every element in $y_{i,j}$ will map to a unique integer. Some set may have finite elements so mapping may be into. If that is not the case and mapping is one to one we know that equivalence exists in any case – Curious May 14 '19 at 6:09
• If $n=2$, then doesn't $f(y_{1,3}) = f(y_{2,1})$? – Anthony Ter May 14 '19 at 6:25