Alexandroff Extension of Rational Number Space is not Hausdorff

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Theorem

Let $\left({\Q, \tau_d}\right)$ be the rational number space under the Euclidean topology $\tau_d$.

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

Let $\Q^* := \Q \cup \left\{{p}\right\}$.

Let $T^* = \left({\Q^*, \tau^*}\right)$ be the Alexandroff extension on $\left({\Q, \tau_d}\right)$.


Then $T^*$ is not a $T_2$ (Hausdorff) space.


Proof

From Condition for Alexandroff Extension to be $T_2$ Space, $T^*$ is a $T_2$ space if and only if $\left({\Q, \tau_d}\right)$ is a locally compact Hausdorff Space.

But from Rational Number Space is not Locally Compact Hausdorff Space, $\left({\Q, \tau_d}\right)$ is not a locally compact Hausdorff Space.

Hence the result.

$\blacksquare$


Sources