Closure of Dense-in-itself is Dense-in-itself in T1 Space

Theorem

Let $T$ be a $T_1$ topological space.

Let $A \subseteq T$.

Let $A$ be dense-in-itself.

Then the closure $A^-$ of $A$ is also dense-in-itself.

Proof

Let $A$ be dense-in-itself.

Then by Dense-in-itself iff Subset of Derivative:

$(1): \quad A \subseteq A'$

where $A'$ denotes the derivative of $A$.

By Derivative of Derivative is Subset of Derivative in $T_1$ Space:

$(2): \quad A \subseteq A'$

By Dense-in-itself iff Subset of Derivative it is sufficient to proof that:

$A^- \subseteq \left({A^-}\right)'$

Thus:

\(\ds \left({A^-}\right)'\)

\(=\)

\(\ds \left({A \cup A'}\right)'\)

Closure Equals Union with Derivative

\(\ds \)

\(=\)

\(\ds A' \cup A\)

Derivative of Union is Union of Derivatives

\(\ds \)

\(=\)

\(\ds A'\)

$(2)$ and Union with Superset is Superset

\(\ds \)

\(=\)

\(\ds A \cup A'\)

$(1)$ and Union with Superset is Superset

\(\ds \)

\(=\)

\(\ds A^-\)

Closure Equals Union with Derivative

$\blacksquare$

Sources

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

• Mizar article TOPGEN_1:36