Quotient Space of Real Line may be Kolmogorov but not Fréchet

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

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


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

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

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

If $p, q \in Y$ and $p ≠ q$, then $p$ or $q$ must be an irrational singleton.

Suppose WLOG that $p$ is an irrational singleton.

Then as shown above, $Y \setminus P$ is open in $Y$, 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 $\{ \Q \}$ is closed in $Y$.

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

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

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

Thus the singleton $\{ \Q \}$ is not closed in $Y$, so $\left({Y, \tau_\sim}\right)$ is not a Fréchet space.