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$\mathbb{R}/\mathbb{Q}$ carries the indiscrete topology, although $\mathbb{R}$ is metrizable and the quotient has more than one point

Statement refuted

Refuted: that a quotient of a metrizable space must be Hausdorff. Here the quotient of $\mathbb{R}$ collapses to the indiscrete topology and, because it has more than one point, is not Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

Witness. Give $\mathbb{R}$ its usual topology (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), identify $\mathbb{Q}$ with its canonical copy in $\mathbb{R}$ (The rationals as equivalence classes of pairs of integers), and set

$x \sim y \quad :\Longleftrightarrow \quad x - y \in \mathbb{Q},$

an equivalence relation. Let $R := \mathbb{R}/\mathbb{Q}$ carry the quotient topology with canonical projection $q$ (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection). Then the only open subsets of $R$ are $\varnothing$ and $R$ : the topology of $R$ is the indiscrete one (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies). Moreover $R$ has more than one point, since $\mathbb{R}$ has an irrational number (The irrationals are uncountable), so this is not the degenerate one-point case.

Facts & Assumptions

Given: $\mathbb{R}$ with its usual topology; the relation $\sim$ ; the quotient $R = \mathbb{R}/\mathbb{Q}$ with projection $q$ ; and a nonempty open $V \subseteq R$ .

[A1]

$q$ is a surjection, $V \subseteq R$ is open exactly when $q^{-1}[V]$ is open in $\mathbb{R}$ , and $q^{-1}[V]$ is saturated: $x \in q^{-1}[V]$ and $y - x \in \mathbb{Q}$ imply $y \in q^{-1}[V]$ (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection).

[A2]

The indiscrete topology on a set is $\{\varnothing, R\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L1]

A nonempty open subset of $\mathbb{R}$ contains a bounded open interval $(a,b)$ with $a < b$ around each of its points (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

[L2]

Strictly between any two reals lies a rational (The rationals embed densely in the reals); equivalently $\mathbb{Q}$ is dense in $\mathbb{R}$ and meets every nonempty open subset (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets).

[L3]

The set of irrationals is uncountable, hence nonempty (The irrationals are uncountable, The rationals as equivalence classes of pairs of integers).

Counterexample

technique · direct

1.1

Let $V \subseteq R$ be open and nonempty, and put $G := q^{-1}[V]$ , which is open in $\mathbb{R}$ by [A1] and nonempty, $q$ being surjective.

2.1

By [L1] there are $a < b$ with $(a,b) \subseteq G$ .

step 1.1L1

3.1

Let $y \in \mathbb{R}$ be arbitrary. By [L2] there is a rational $t$ with $a - y < t < b - y$ , so $y + t \in (a,b)$ .

step 2.1L2

4.1

By step 2.1 and step 3.1 the point $y + t$ lies in $G$ , and $(y+t) - y = t \in \mathbb{Q}$ , so $y \in G$ by the saturation clause of [A1]. As $y$ was arbitrary, $G = \mathbb{R}$ and hence $V = q[G] = R$ , $q$ being surjective.

step 2.1step 3.1A1

5.1

So the only open subsets of $R$ are $\varnothing$ and $R$ , which is the indiscrete topology by [A2].

step 4.1A2

6.1

By [L3] there is an irrational $\alpha$ , and $\alpha - 0 = \alpha \notin \mathbb{Q}$ , so $q(\alpha) \ne q(0)$ and $R$ has at least two points; with step 5.1 the quotient of the metrizable space $\mathbb{R}$ is a space with more than one point carrying the indiscrete topology, which is not Hausdorff, no two distinct points having disjoint open neighbourhoods.

step 5.1A2L3∎

Remarks

What has been destroyed and what has not. The quotient still has more than one point, as step 6.1 proves. What is destroyed is every proper nonempty open set: a saturated open set is a union of cosets, each of which is dense, so a saturated open set that is nonempty is everything.
This is the extreme case of the failure recorded on the general page. The line with two origins (Two copies of $\mathbb{R}$ glued along $\mathbb{R} \setminus \{0\}$ give a non-Hausdorff quotient of a metrizable space, by an open quotient map) loses the Hausdorff condition at exactly two points; here the quotient topology retains nothing at all. Both quotient maps are open, so openness of the quotient map is no protection whatever.
The quotient map is open here too. For open $U \subseteq \mathbb{R}$ the saturation is $\bigcup_{t \in \mathbb{Q}} (U + t)$ , a union of translates and hence open, exactly as in $\mathbb{R}/\mathbb{Z}$ : the quotient map is open, and the quotient is homeomorphic to $[0,1]$ with its endpoints identified; step 4.1 shows that this union is $\mathbb{R}$ whenever $U$ is nonempty, which is the whole phenomenon in one line.

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Sources

Quotient space (topology) (Wikipedia) (standard reference, not scraped)
Trivial topology (Wikipedia) (standard reference, not scraped)
Topology, Spring 2005, Homework 1 [Flagg/Blecher] (standard reference, not scraped)

R/Q\mathbb{R}/\mathbb{Q}R/Q carries the indiscrete topology, although R\mathbb{R}R is metrizable and the quotient has more than one point