Quotient Space of Real Line may be Kolmogorov but not Fréchet

Theorem
Let $T = \struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.

Let $\sim$ be an equivalence relation on $\R$.

Let $\struct {\R / {\sim}, \tau_\sim}$ be the quotient space of $\R$ by $\sim$.

Then it is possible for $\struct {\R / {\sim}, \tau_\sim}$ to be a Kolmogorov space but not a Fréchet space.

Proof by Counterexample
Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.

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


 * $x = y$

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

Let $\struct {\R / {\sim}, \tau_\sim}$ be the quotient space of $\R$ by $\sim$.

Then $\struct {\R / {\sim}, \tau_\sim}$ is a Kolmogorov space but not a Fréchet space.

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

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

Note that:


 * $\map \phi x = \set x$ if $x$ is irrational.
 * $\map \phi x = \Q$ if $x$ is rational.

Kolmogorov
Let $x$ be irrational.

Then:
 * $\phi^{-1} \sqbrk {Y \setminus \set x} = \R \setminus \set x$

Thus $Y \setminus \set x$ is open in $Y$.

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

Then $\set p$ or $\set q$ must be a singleton containing an irrational number.

, suppose that $\set p$ 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
$\set \Q$ is closed in $Y$.

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

Thus $\phi^{-1} \sqbrk {\set \Q} = \Q$ is closed in $\R$.

But this contradicts the fact that $\Q \subsetneqq \R$ and Rationals are Everywhere Dense in Topological Space of Reals.

Thus the singleton $\set \Q$ is not closed in $Y$.

Hence $\struct {Y, \tau_\sim}$ is not a Fréchet space.