Topological Space induced by Neighborhood Space induced by Topological Space

Theorem

Let $S$ be a set.

Let $\tau$ be a topology on $S$, thus forming the topological space $\struct {S, \tau}$.

Let $\struct {S, \NN}$ be the neighborhood space induced by $\tau$ on $S$.

Let $\struct {S, \tau'}$ be the topological space induced by $\NN$ on $S$.

Then $\tau = \tau'$.

Proof

Let $U \in \tau$ be an open set of $\struct {S, \tau}$.

By Set is Open iff Neighborhood of all its Points, $U$ is a neighborhood of each of its points.

By definition, $U$ is an open set of $\struct {S, \NN}$.

Thus by definition of neighborhood space induced by $\tau$ on $S$:

$U \in \NN$

Then, by definition of the topological space induced by $\NN$ on $S$:

$U \in \tau'$.

Thus $\tau \subseteq \tau'$.

Now let $U \in \tau'$ be an open set of $\struct {S, \tau'}$.

Then, by definition, in the neighborhood space $\struct {S, \NN}$, $U$ is an open set of $\struct {S, \NN}$.

That is, $U$ is a neighborhood in $\struct {S, \NN}$ of each of its points.

But the neighborhoods of $\struct {S, \NN}$ are the neighborhoods of $\struct {S, \tau}$.

Thus by Set is Open iff Neighborhood of all its Points, $U$ is an open set of $\struct {S, \tau}$.

That is:

$U \in \tau$.

It follows by definition of set equality that $\tau = \tau'$.

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $3$: Topological Spaces: $\S 3$: Neighborhoods and Neighborhood Spaces