Quotient Space of Real Line may be Kolmogorov but not Fréchet

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

Define an equivalence relation $\sim$ by letting $x \sim y$ either:


 * $x = y$

or:
 * $x, y \in \Q$

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

Then $\left({\R / {\sim}, \tau_\sim}\right)$ is a Kolmogorov space but not a Fréchet space.

Proof
Let $Y = \R / {\sim}$.

Let $\phi: \R \to Y$ be the quotient mapping.

Note that:


 * $\phi \left({x}\right) = \left\{{x}\right\}$ if $x$ is irrational.
 * $\phi \left({x}\right) = \Q$ if $x$ is rational.

Kolmogorov
If $x$ is irrational, then $\phi^{-1} \left({Y \setminus \left\{{x}\right\}}\right) = \R \setminus \left\{{x}\right\}$.

Thus $Y \setminus \left\{{x}\right\}$ is open in $Y$.

Let $p, q \in Y$ such that $p \ne q$.

Then $\left\{{p}\right\}$ or $\left\{{g}\right\}$ must be a singleton containing an irrational number.

Suppose WLOG that $\left\{{p}\right\}$ is a singleton containing an irrational number.

Then as shown above, $Y \setminus P$ is open in $Y$.

Thus so $p$ and $q$ are distinguishable.

Since this holds for any two points in $Y$, the space is Kolmogorov.

Not Fréchet
Suppose for the sake of contradiction that $\left\{{\Q}\right\}$ is closed in $Y$.

By Identification Mapping is Continuous, $\phi$ is continuous.

Thus $\phi^{-1}\left({\left\{{\Q}\right\}}\right) = \Q$ is closed in $\R$.

But this contradicts the fact that $\Q \subsetneqq \R$ and Rationals Dense in Reals.

Thus the singleton $\left\{{\Q}\right\}$ is not closed in $Y$.

Hence $\left({Y, \tau_\sim}\right)$ is not a Fréchet space.