Alexandroff Extension of Rational Number Space is Connected

From ProofWiki
Jump to navigation Jump to search


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 connected space.


By definition $T^*$ is a connected space of $T^*$ if and only if it admits no separation.

Aiming for a contradiction, suppose $T^*$ does admit a separation.

That is, there exist open sets $A, B \in \tau^*$ such that $A, B \ne \O$, $A \cup B = \Q^*$ and $A \cap B = \O$.

That is, both $A$ and $B = \relcomp {\Q^*} A$ are open in $T^*$.

Without loss of generality, Let $p \in A$.

Then $B$ is a closed and compact subset of $\Q$.

From Compact Set of Rational Numbers is Nowhere Dense, $B$ is nowhere dense.

From Set is Closed iff Equals Topological Closure, $B$ equals its closure.

Thus by definition of nowhere dense, the interior of $B$ is empty.

We have that $B$ is open in $T^*$ and $p \notin B$.

Thus it follows by definition of Alexandroff extension that $B$ is open in $\struct {\Q, \tau_d}$.

But from Interior of Open Set, $B$ equals its interior.

This contradicts the deduction above that the interior of $B$ is empty.

Hence by Proof by Contradiction $B$ cannot be open in $T^*$.

That is, $A$ and $B$ cannot form a separation of $T^*$.

Thus $T^*$ is a connected space by definition.