# Equivalence of Definitions of Topology Generated by Synthetic Basis

## Theorem

Let $S$ be a set.

Let $\mathcal B$ be a synthetic basis on $S$.

Let $\tau$ be the topology on $S$ generated by the synthetic basis $\mathcal B$:

$\displaystyle \tau = \left\{{\bigcup \mathcal A: \mathcal A \subseteq \mathcal B}\right\}$

Then:

$(1): \quad \forall U \subseteq S: U \in \tau \iff U = \displaystyle \bigcup \left\{ {B \in \mathcal B: B \subseteq U}\right\}$
$(2): \quad \forall U \subseteq S: U \in \tau \iff \forall x \in U: \exists B \in \mathcal B: x \in B \subseteq U$

## Proof

### Proof of $(1)$

Trivially, the reverse implication holds, as $\left\{{B \in \mathcal B: B \subseteq U}\right\} \subseteq \mathcal B$.

We now show that the forward implication holds.

Suppose that $U \in \tau$. Then, by definition:

$\displaystyle \exists \mathcal A \subseteq \mathcal B: U = \bigcup \mathcal A$

By Union is Smallest Superset: General Result, we have::

$\displaystyle \forall B \in \mathcal A: B \subseteq U$

By the definition of a subset, it follows that:

$\displaystyle \mathcal A \subseteq \left\{{B \in \mathcal B: B \subseteq U}\right\}$

Therefore:

$\displaystyle U = \bigcup \mathcal A \subseteq \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\}$

$\blacksquare$

Definition of a Subset:

$\displaystyle \mathbb A \subseteq \mathbb B \iff \forall \mathcal t \left[ \mathcal t \in \mathbb A \implies \mathcal t \in \mathbb B \right]$

Definition of a Generalized Union:

$\displaystyle \mathcal t \in \bigcup_{}^{} \left[ \mathbb x \right] \iff \exists \mathcal q \left[ \mathcal q \in \mathbb x \wedge \mathcal t \in \mathcal q \right]$

The following is a tautology:

$\displaystyle \left[ \mathcal t \in \mathbb A \implies \mathcal t \in \mathbb B \right] \implies \left[ \left( \mathcal t \in \mathbb A \wedge \mathcal q \in \mathcal t \right) \implies \left( \mathcal t \in \mathbb B \wedge \mathcal q \in \mathcal t \right) \right]$

Bind "t" with a universal quantifier:

$\displaystyle \forall \mathcal t \left[ \left[ \mathcal t \in \mathbb A \implies \mathcal t \in \mathbb B \right] \implies \left[ \left( \mathcal t \in \mathbb A \wedge \mathcal q \in \mathcal t \right) \implies \left( \mathcal t \in \mathbb B \wedge \mathcal q \in \mathcal t \right) \right] \right]$

Distribute the universal quantifier and utilize the definition for a subset:

$\displaystyle \forall \mathcal t \left[ \mathcal t \in \mathbb A \implies \mathcal t \in \mathbb B \right] \implies \forall \mathcal t \left[ \left( \mathcal t \in \mathbb A \wedge \mathcal q \in \mathcal t \right) \implies \left( \mathcal t \in \mathbb B \wedge \mathcal q \in \mathcal t \right) \right]$
$\displaystyle \mathbb A \subseteq \mathbb B \implies \forall \mathcal t \left[ \left( \mathcal t \in \mathbb A \wedge \mathcal q \in \mathcal t \right) \implies \left( \mathcal t \in \mathbb B \wedge \mathcal q \in \mathcal t \right) \right]$

The operation used to distribute the universal quantifier looks as follows:

$\displaystyle \forall \mathcal x \left[ \theta (x) \to \phi (x) \right] \implies \left( \forall \mathcal x \left[ \theta (x) \right] \to \forall \mathcal x \left[ \phi (x) \right] \right)$

Utilizing contraposition and replacing, respectively, "phi" with "not alpha" and "theta" with "not beta" makes the following valid formula scheme:

$\displaystyle \forall \mathcal x \left[ \alpha (x) \to \beta (x) \right] \implies \left( \exists \mathcal x \left[ \alpha (x) \right] \to \exists \mathcal x \left[ \beta (x) \right] \right)$

So far there are two formulas that can derive, via modus ponens, the following formula:

$\displaystyle \mathbb A \subseteq \mathbb B \implies \mathbf{\forall \mathcal t \left[ \left( \mathcal t \in \mathbb A \wedge \mathcal q \in \mathcal t \right) \implies \left( \mathcal t \in \mathbb B \wedge \mathcal q \in \mathcal t \right) \right]}$
$\displaystyle \mathbf{\forall \mathcal t \left[ \left( \mathcal t \in \mathbb A \wedge \mathcal q \in \mathcal t \right) \implies \left( \mathcal t \in \mathbb B \wedge \mathcal q \in \mathcal t \right) \right]} \implies \left( \exists \mathcal t \left[ \mathcal t \in \mathbb A \wedge \mathcal q \in \mathcal t \right] \implies \exists \mathcal t \left[ \mathcal t \in \mathbb B \wedge \mathcal q \in \mathcal t \right] \right)$
$\displaystyle \therefore \mathbb A \subseteq \mathbb B \implies \left( \exists \mathcal t \left[ \mathcal t \in \mathbb A \wedge \mathcal q \in \mathcal t \right] \implies \exists \mathcal t \left[ \mathcal t \in \mathbb B \wedge \mathcal q \in \mathcal t \right] \right)$

This can be generalized for "q":

$\displaystyle \forall \mathcal q \left[ \mathbb A \subseteq \mathbb B \implies \left( \exists \mathcal t \left[ \mathcal t \in \mathbb A \wedge \mathcal q \in \mathcal t \right] \implies \exists \mathcal t \left[ \mathcal t \in \mathbb B \wedge \mathcal q \in \mathcal t \right] \right) \right]$

The following is a valid formula scheme:

$\displaystyle \forall \mathcal q \left[ \phi (x) \to \theta (q) \right] \implies \left( \phi (x) \to \forall \mathcal q \left[ \theta (q) \right] \right)$

Thus, one can utilize this scheme as an operation and utilize it on the generalized formula:

$\displaystyle \mathbb A \subseteq \mathbb B \implies \forall \mathcal q \left( \exists \mathcal t \left[ \mathcal t \in \mathbb A \wedge \mathcal q \in \mathcal t \right] \implies \exists \mathcal t \left[ \mathcal t \in \mathbb B \wedge \mathcal q \in \mathcal t \right] \right)$

Now, after manipulating the first two definitions proposed into the previous formula yields:

$\displaystyle \mathbb A \subseteq \mathbb B \implies \bigcup_{}^{} \mathbb A \subseteq \bigcup_{}^{} \mathbb B$

$\blacksquare$

$\displaystyle \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\} \subseteq U$

By definition of set equality:

$\displaystyle U = \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\}$

$\Box$

### Proof of $(2)$

From Set is Subset of Union: General Result, the forward implication directly follows.

We now show that the reverse implication holds.

By hypothesis, we have that:

$\displaystyle U \subseteq \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\}$
$\displaystyle \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\} \subseteq U$

By definition of set equality:

$\displaystyle U = \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\}$

The result follows.

$\blacksquare$