Characterization of Closure by Basis

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $\BB \subseteq \tau$ be a basis.

Let $A$ be a subset of $S$.

Let $x$ be a point of $S$.

Then $x \in A^-$ if and only if:

$\forall U \in \BB: x \in U \implies A \cap U \ne \O$

where:

$A^-$ denotes the closure of $A$

Proof

Sufficient Condition

Let $x \in A^-$.

Let $U \in \BB$.

By definition of basis, $U$ is an open set of $T$.

Thus from Condition for Point being in Closure:

if $x \in U$ then $A \cap U \ne \O$.

$\Box$

Necessary Condition

Let $x$ be such that for every $U \in \BB$:

if $x \in U$

then $A \cap U \ne \O$.

By Condition for Point being in Closure, to prove that $x \in \operatorname{Fr} A$ it is enough to prove that:

for every open set $U$ of $T$:

if $x \in U$ then $A \cap U \ne \O$.

Let $U$ be an open set of $T$.

Let $x \in U$.

By definition of (analytic) basis, there exists $V \in \BB$ such that:

$x \in V \subseteq U$

By assumption:

$A \cap V \ne \O$

From the corollary to Set Intersection Preserves Subsets:

$A \cap V \subseteq A \cap U$

So:

$A \cap U \ne \O$

and hence the result.

$\blacksquare$

Sources

• 1989: Ryszard Engelking: General Topology (revised and completed ed.)

• Mizar article YELLOW_9:37