Prove that a collection $\mathcal{S}$ of subsets $U$ of an infinite set $X$, $|X \backslash U|=n$, is a subbasis for the cofinite topology on $X$. I'm a non-mathematician who's taken an interest in math only recently. I've been studying real analysis and linear algebra by myself for the past couple of months. I thought I'd explore some other areas of mathematics and have been trying my hand at topology for the past couple of weeks. I came across this question on an assignment in a MOOC I've enrolled in:
Prove that a collection $\mathcal{S}$ of subsets $U$ of an infinite set $X$, $|X \backslash U|=n$, is a subbasis for the cofinite topology on $X$.
This is the proof I wrote:
$co\mathcal{F}:=\{U \subset X: |X\backslash U| = i, i$ varying in $\mathbb N\}.$ For any fixed $n\in \mathbb N$, let $\mathcal{S}=\{U\subset X: |X\backslash U|=n\}.$ Now, $\mathcal{S}\subset co\mathcal{F}.$ We shall prove that $\mathcal{S}$ is a basis for the cofinite topology, and thereby, a subbasis.
We require that $\mathcal{S}$ meets two conditions:
1: $\bigcup_{U \in \mathcal{S}}U=X.$
2: For $U_1, U_2 \in \mathcal{S}$, and for $x \in U_1\cap U_2, \exists \ U_3\in \mathcal{S}$, such that $x\in U_3\subseteq U_1\cap U_2.$
Proof of 1:
Suppose $\bigcup_{U \in \mathcal{S}}U\neq X.$ Then $\exists \ \mathcal{S'}$ such that $\mathcal{S'}\cap\mathcal{S}=\emptyset$, and $\mathcal{S}\cup\mathcal{S'}=X.$ Now, $\mathcal{S'}$ must be finite in $X$, owing to the fact that $\mathcal{S'}=X\backslash \mathcal{S}$. Then $\mathcal{S'}$ contains only finite subsets, and $\mathcal{S}$ contains all the cofinite subsets of X. $\implies \mathcal{S}=co\mathcal{F}.$ $X\subset co\mathcal{F} \implies X\subset \mathcal{S}.$ Thus, we have $\mathcal{S}=X$, which is a contradiction. Therefore, I holds.
Proof of 2:
Let $U_3:= U_1\cap U_2 \neq \emptyset.$ Since $U_1$ and $U_2 \in co\mathcal{F}$, their intersection $U_3$ is also in $co\mathcal{F}.$ Therefore, $U_3\subset X.$ Now, it will suffice to show that $|X\backslash U_3|=n$. Suppose not. Then $U_3 \notin \mathcal{S}.$ But $X=\bigcup_{U\in\mathcal{S}}U, \implies U_3\in \bigcup_{U\in\mathcal{S}}U$, which is a contradiction. Therefore, $|X\backslash U_3|=n$, and $\mathcal{S}$ forms a basis, and thereby a subbasis, for the cofinite topology on X.
Is my line of reasoning correct?
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NB: This is my first question here, so I'm sorry if there are any errors/omitted details.
 A: As an example : let $X$ be infinite and $\mathcal{S}=\{U \subseteq X\mid |S\setminus U|=3\}$ be a collection of subsets and $\mathcal{T}$ is the smallest topology that contains $\mathcal{S}$. We need to prove $\mathcal{T}=\mathcal{T}_{cf}$, the cofinite topology on $X$. It\s clear that as all members of $\mathcal{S}$ are cofinite, by minimality $\mathcal{T}\subseteq \mathcal{T}_{cf}$.
If $F$ is a finite subset of size $4$, say $F=\{x_1,x_2,x_3,x_4\}$, then note that $$X\setminus F = (X\setminus \{x_1,x_2,x_3\}) \cap (X\setminus \{x_1,x_2,x_4\}$$ The latter intersection is in $\mathcal{T}$ as it's an intersection of two sets from $\mathcal{S}$. It follows that any complement of a finite set of size $4$ is in $\mathcal{T}$. Find similar arguments for complements of sets of any finite size $>3$.
And if $F = \{x_1,x_2\}$ we can find distinct $x_3,x_4,x_5,x_6 \notin F$ and write $$X\setminus F = (X\setminus \{x_1,x_2, x_3,x_4\}) \cup (X\setminus \{x_1,x_2,x_5,x_6\}) \in \mathcal{T}$$ as topologies are closed under unions. So we also have complements of doubletons (and singletons; try that too) in $\mathcal{T}$. So finally $\mathcal{T}_{cf} \subseteq \mathcal{T}$ for this case (sketched), and we're done.
Try to generalise this argument to all $n$.
