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 and let $\mathbb I$ be the irrational numbers.

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

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

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 $x \in \Q$, $\phi(x) = \Q$, and for $x \in \mathbb I$, $\phi(x) = \mathbb I$.

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

Then by the definition of the quotient topology:


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

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

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

Then $\varnothing \subsetneqq \Q = \phi^{-1}( \{ \Q \} ) \in \tau$.

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 $\{ \Q \}$ and $\{ \mathbb I \}$.

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