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

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: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ 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

$x\sim y:⟺x-y\in \mathbb{Q},$

an equivalence relation. Let $R:=\mathbb{R}\mathrm{/}\mathbb{Q}$ 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 $\mathrm{\varnothing }$ 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.

• 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 $\mathbb{R}\mathrm{/}\mathbb{Z}$: 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.