# Generated Topology is a Topology

This article has been proposed for deletion. In particular: duplicate of Synthetic Basis formed from Synthetic Sub-Basis |

## Theorem

Let $X$ be a set.

Let $\mathcal S \subseteq \mathcal P \left({X}\right)$, where $\mathcal P \left({X}\right)$ is the power set of $X$.

Let $\mathcal T_\mathcal S$ be the generated topology for $\mathcal S$.

Then $\mathcal T_\mathcal S$ is a topology on $X$.

## Proof

We show that $\mathcal S^* = \left\{{\bigcap S : S \subseteq \mathcal S \text{ finite}}\right\}$ (cf. the definition of the generated topology) is a basis.

To see that $\mathcal S^*$ is a basis, we need to prove two things:

- $(1): \quad X = \bigcup \mathcal S^*$
- $(2): \quad$ For any $U_1, U_2 \in \mathcal S^*$ and $x \in U_1 \cap U_2$ there is a $U \in \mathcal S^*$ such that $x \in U \subseteq U_1 \cap U_2$

First note that $X = \bigcap \varnothing \in \mathcal S^*$ and therefore $\bigcup \mathcal S^* = X$.

Additionally, if $U_1, U_2 \in \mathcal S^*$, then there exist finite sets $S_1, S_2 \subseteq \mathcal S$ such that $U_1 = \bigcap S_1$ and $U_2 = \bigcap S_2$ by the definition of $\mathcal S^*$.

Thus we have:

- $U_1 \cap U_2 = \bigcap (S_1 \cup S_2)$

Because $S_1 \cup S_2$ is again a finite set it follows that $U_1 \cap U_2 \in \mathcal S^*$.

This implies $(2)$.

$\blacksquare$