Indiscrete Topology is Topology

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Theorem

Let $S$ be a set.

Let $\tau$ be the indiscrete topology on $S$.

$\tau$ is a topology on $S$.


Proof

Let $T = \left({S, \left\{{\varnothing, S}\right\}}\right)$ be the indiscrete space on $S$.


Confirming the criteria for $T$ to be a topology:

$(1): \quad$ Trivially, by definition, $\varnothing \in \tau$ and $S \in \tau$.
$(2): \quad \varnothing \cup \varnothing = \varnothing \in \tau$, $\varnothing \cup S = S \in \tau$ and $S \cup S = S \in \tau$ from Union with Empty Set and Union is Idempotent.
$(3): \quad \varnothing \cap \varnothing = \varnothing \in \tau$, $\varnothing \cap S = \varnothing \in \tau$ and $S \cap S = S \in \tau$ from Intersection with Empty Set and Intersection is Idempotent.

$\blacksquare$


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