Quotient Space of Real Line may be Indiscrete

Theorem
Let $T = \left({\R, \tau}\right)$ be the real numbers with the usual topology.

Let $\Q$ be the rational numbers.

Let $\mathbb I$ be the irrational numbers.

Then $\left\{ {\Q, \mathbb I}\right\}$ is a partition of $\R$.

Let $\sim$ be the equivalence relation induced on $\R$ by $\left\{ {\Q, \mathbb I}\right\}$.

Let $T_\sim := \left({\R / {\sim}, \tau_\sim}\right)$ be the quotient space of $\R$ by $\sim$.

Then $T_\sim$ is an indiscrete space.

Proof
Let $\phi: \R \to \R/{\sim}$ be the quotient mapping.

Then:
 * for $x \in \Q$, $\phi \left({x}\right) = \Q$
 * for $x \in \mathbb I$, $\phi \left({x}\right) = \mathbb I$.

Suppose for the sake of contradiction that $\left\{ {\mathbb I}\right\} \in \tau_\sim$.

Then by the definition of the quotient topology:


 * $\varnothing \subsetneqq \mathbb I = \phi^{-1} \left({\left\{{\mathbb I}\right\} }\right) \in \tau$

Thus by Rationals Dense in Reals, $\mathbb I$ contains a rational number, a contradiction.

Suppose for the sake of contradiction that $\left\{ {\Q}\right\} \in \tau_\sim$.

Then:
 * $\varnothing \subsetneqq \Q = \phi^{-1} \left({\left\{{\Q}\right\} }\right) \in \tau$

Thus by Irrationals Dense in Reals, $\Q$ contains a irrational number, a contradiction.

As $\R / {\sim}$ has exactly two elements, its only non-empty proper subsets are $\left\{ {\Q}\right\}$ and $\left\{ {\mathbb I}\right\}$.

As neither of these sets is $\tau_\sim$-open, $\left({\R / {\sim}, \tau_\sim}\right)$ is indiscrete.