Modified Fort Topology is Topology
Theorem
Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $\tau_{a, b}$ is a topology on $S$.
Proof
Recall the definition of a modified Fort space:
Let $N$ be an infinite set.
Let $\set a$ and $\set b$ be singleton sets such that $a \ne b$ and $a, b \notin N$.
Let $S = N \cup \set a \cup \set b$.
Let $\tau_{a, b}$ be the set of subsets of $S$ defined as:
- $\tau_{a, b} = \set {H \subseteq N} \cup \set {H \subseteq S: \paren {a \in H \lor b \in H} \land N \setminus H \text { is finite} }$
That is, a subset $H$ of $S$ is in $\tau_{a, b}$ if and only if either:
- $(1): \quad H$ is any subset of $N$
or:
- $(2): \quad$ if $a$ or $b$ or both are in $H$, then $H$ is in $S$ only if it is cofinite in $S$, that is, that it contains all but a finite number of points of $S$ (or $N$, equivalently).
Then $\tau_{a, b}$ is a modified Fort topology on $a$ and $b$, and the topological space $T = \struct {S, \tau_{a, b} }$ is a modified Fort space.
Let $S = N \cup \set {a, b}$ where $N$ is infinite, $a \ne b$ and $a, b \notin N$.
We have that $\O \subseteq N$ so $\O \in \tau_{a, b}$.
We have that $a, b \in S, S \setminus S = \O$ and $\O$ is trivially finite, so $S \in \tau_{a, b}$.
Now consider $A, B \in \tau_{a, b}$, and let $H = A \cap B$.
Let either $A \subseteq N$ or $B \subseteq N$.
From Intersection is Subset we have that $A \cap B \subseteq N$ and $A \cap B \subseteq B$.
Hence $A \cap B \subseteq N$ from Subset Relation is Transitive.
So by definition $A \cap B \in \tau_{a, b}$
Now suppose $A \cap \set {a, b} \ne \O$ and $B \cap \set {a, b} \ne \O$.
Then:
\(\ds H\) | \(=\) | \(\ds A \cap B\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds N \setminus H\) | \(=\) | \(\ds N \setminus \paren {A \cap B}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {N \setminus A} \cup \paren {N \setminus B}\) | De Morgan's Laws: Difference with Intersection |
In order for $A$ and $B$ to be open sets we have that $N \setminus A$ and $N \setminus B$ are both finite.
Hence their union is also finite and so $N \setminus \paren {A \cap B}$ is finite.
So $H = A \cap B \in \tau_{a, b}$ as its complement is finite.
Now let $\UU \subseteq \tau_{a, b}$.
Then from De Morgan's Laws: Difference with Union:
- $\ds N \setminus \paren {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \paren {N \setminus U}$
We have either of two options:
- $(1): \quad \forall U \in \UU: U \subseteq N$
in which case:
- $\ds \bigcup \UU \subseteq N$
Or:
- $(2): \quad \exists U \in \UU: N \setminus U$ is finite
in which case:
- $\ds \bigcap_{U \mathop \in \UU} \paren {N \setminus U}$ is finite, from Intersection is Subset.
So in either case:
- $\ds \bigcup \UU \in \tau_{a, b}$
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $27$. Modified Fort Space