Rational Numbers are Totally Disconnected/Proof 1

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Then $\struct {\Q, \tau_d}$ is a totally disconnected space.


Proof

Let $x, y \in \Q$ such that $x \ne y$.

From Between two Rational Numbers exists Irrational Number, there exists $\alpha \in \R \setminus \Q$ such that $x < \alpha < y$.

From Rational Numbers are not Connected, it follows that $x$ and $y$ belong to different components of $\Q$.

As $x$ and $y$ are arbitrary, it follows that no rational number is in the same component as any other rational number.

That is, the components of $\Q$ are singetons.

Hence the result, by definition of a totally disconnected space.

$\blacksquare$