Finite Complement Space is Irreducible
Jump to navigation
Jump to search
Theorem
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then $T$ is irreducible.
Proof
Let $U_1, U_2 \in \tau$ be non-empty open sets of $T$.
By definition, $S$ is infinite.
By definition, $\relcomp S {U_1}$ and $\relcomp S {U_2}$ are finite.
From Complement of Finite Subset of Infinite Set is Infinite, it follows that $U_1$ and $U_2$ are both infinite.
From Infinite Subset of Finite Complement Space Intersects Open Sets:
- $U_1 \cap U_2\ne \O$
That is, $U_1$ and $U_2$ intersect each other.
As $U_1$ and $U_2$ are arbitrary, the result follows by definition of irreducible space.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $6$