Alexandroff Extension of Rational Number Space is T1 Space

Theorem

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.

Let $p$ be a new element not in $\Q$.

Let $\Q^* := \Q \cup \set p$.

Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\struct {\Q, \tau_d}$.

Then $T^*$ is a $T_1$ (Fréchet) space.

Proof

From Condition for Alexandroff Extension to be $T_1$ Space, $T^*$ is a $T_1$ space if and only if $\struct {\Q, \tau_d}$ is also a $T_1$ space.

From Rational Numbers form Metric Space, $\struct {\Q, d}$ is a metric space.

From Metric Space fulfils all Separation Axioms, $\struct {\Q, \tau_d}$ is a $T_1$ space.

Hence the result.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $35$. One Point Compactification Topology: $4$