Quotient Space of Real Line may be Kolmogorov but not Fréchet

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Theorem

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

Define an equivalence relation $\sim$ by letting $x \sim y$ if and only if 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.


Proof

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

If $x$ is 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.

Without loss of generality, 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.

$\Box$


Not Fréchet

Aiming for a contradiction, suppose $\set \Q$ 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.

$\blacksquare$