Characterization of Fineness of Topology in terms of Topological Bases

Theorem

Let $X$ be a set.

Let $\tau$ and $\tau'$ be topologies on $X$.

Let $\BB \subseteq \tau$ be a basis for $\tau$ and $\BB' \subseteq \tau'$ be a basis for $\tau'$.

The following statements are equivalent:

$(1): \quad$ $\tau'$ is finer than $\tau$

$(2): \quad$ for each $x \in X$ and each $B \in \BB$ containing $x$, there exists $B' \in \BB'$ such that $x \in B' \subseteq B$.

Proof

$(1)$ implies $(2)$

Let $x \in X$ and $B \in \BB$ contain $x$.

Since $\tau'$ is finer than $\tau$, we have that $B \in \tau'$.

Since $\BB'$ is a basis for $\tau'$, there exists a subset $\set {B'_\alpha : \alpha \in I} \subseteq \BB'$ such that:

$\ds B = \bigcup_{\alpha \mathop \in I} B'_\alpha$

from Open Set is Union of Elements of Basis.

Since $x \in B$, we have $x \in B'_\alpha$ for some $\alpha \in I$.

We have $B'_\alpha \subseteq B$ from Set is Subset of Union.

Setting $B = B'_\alpha$, we have therefore found $B' \in \BB'$ such that $x \in B' \subseteq B$, proving $(2)$.

$\Box$

$(2)$ implies $(1)$

Let $U \in \tau$.

We aim to prove that $U \in \tau'$.

Let $x \in U$.

Since $\BB$ is a basis for $\tau$, there exists a subset $\set {B_\alpha : \alpha \in I} \subseteq \BB$ such that:

$\ds U = \bigcup_{\alpha \mathop \in I} B_\alpha$

from Open Set is Union of Elements of Basis.

Then for some $\alpha \in I$, we have $x \in B_\alpha \subseteq U$.

By $(2)$, it follows that there exists $B' \in \BB'$ such that:

$x \in B' \subseteq B_\alpha$

in particular:

$x \in B' \subseteq U$

We have now shown that for each $x \in U$, there exists $B'_x \in \BB'$ such that:

$x \in B'_x \subseteq U$

We argue that:

$\ds U = \bigcup_{x \in U} B'_x$

Clearly:

$\ds U \subseteq \bigcup_{x \in U} B'_x$

since $x \in B'_x$ for each $x \in U$, while:

$\ds \bigcup_{x \in U} B'_x \subseteq U$

sine $B'_x \subseteq U$ for each $x \in U$.

So we have:

$\ds U = \bigcup_{x \in U} B'_x$

So $U$ is the union of elements of $\BB'$.

In particular, $U$ is the union of elements of $\tau'$.

So, since topologies are closed under union, we have $U \in \tau'$.

So we have $\tau \subseteq \tau'$ as desired.

$\blacksquare$