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FALSE: a quotient of a Hausdorff space is Hausdorff

Statement

False claim: if $X$ is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) and $q : X \to Y$ is a quotient map (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 $Y$ is Hausdorff.

The refutation is the line with two origins. Let

$S \;:=\; \mathbb{R} \sqcup \mathbb{R} \;=\; \bigsqcup_{i < 2} \mathbb{R}$

be the disjoint union of two copies of $\mathbb{R}$ with its usual topology (The disjoint union (coproduct) $\bigsqcup_i X_i$ with the final topology of the canonical injections: a set is open exactly when each of its traces is, 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), whose points are the pairs $(x,i)$ with $x \in \mathbb{R}$ and $i < 2$ . Let $\sim$ be the equivalence relation on $S$ whose classes are

$\{(x,0), (x,1)\} \ \ (x \ne 0), \qquad \{(0,0)\}, \qquad \{(0,1)\} ,$

and let $L := S/\!\sim$ with the quotient topology and canonical projection $q$ . Then $S$ is Hausdorff and $L$ is not: the two classes $q(0,0)$ and $q(0,1)$ , the "two origins", cannot be separated by disjoint open sets.

Facts & Assumptions

Given: The space $S = \bigsqcup_{i<2}\mathbb{R}$ with the disjoint union topology, the relation $\sim$ above, the quotient $L = S/\!\sim$ with its canonical projection $q$ , and the two points $a := q(0,0)$ and $b := q(0,1)$ of $L$ .

[A1]

$U \subseteq S$ is open exactly when both traces $U_i = \{\, x \in \mathbb{R} : (x,i) \in U \,\}$ are open in $\mathbb{R}$ ; each set $\mathbb{R} \times \{i\}$ is open in $S$ ; and $\kappa_i[V] = V \times \{i\}$ is open in $S$ whenever $V$ is open in $\mathbb{R}$ (The disjoint union (coproduct) $\bigsqcup_i X_i$ with the final topology of the canonical injections: a set is open exactly when each of its traces is, A map out of a disjoint union is continuous iff each of its restrictions is; the canonical injections are open and closed embeddings; and each summand is clopen in the union).

[A2]

The classes listed in the statement are pairwise disjoint and cover $S$ , so $\sim$ is an equivalence relation; $q$ is a surjection and $V \subseteq L$ is open exactly when $q^{-1}[V]$ is open in $S$ (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection).

[A3]

A space is Hausdorff when distinct points have disjoint open neighbourhoods (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L1]

$(a,b) = \{t : a < t < b\}$ is open in the usual topology of $\mathbb{R}$ ; a set is open there exactly when each of its points has a bounded open interval around it inside the set; and $a < (a+b)/2 < b$ whenever $a < b$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, 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).

[L2]

The order of $\mathbb{R}$ is total, so a two-element set of reals has a minimum, which lies in the set and is a lower bound for it (Maximum and minimum of a set); $|s| \ge 0$ , and $|s| > 0$ when $s \ne 0$ (Basic properties of the absolute value); and $|s - u| \le |s - t| + |t - u|$ (The triangle inequality).

Refutation

technique · direct

1.1

$S$ is Hausdorff. Let $(x,i) \ne (y,j)$ in $S$ . If $i \ne j$ then $\mathbb{R} \times \{i\}$ and $\mathbb{R} \times \{j\}$ are disjoint open sets containing them, by [A1]. If $i = j$ then $x \ne y$ ; put $r := |x-y|/2 > 0$ by [L2], and take $(x-r,x+r) \times \{i\}$ and $(y-r,y+r) \times \{i\}$ , which are open by [A1] and [L1] and are disjoint, since a common point $t$ would give $|x-y| \le |x-t| + |t-y| < 2r = |x-y|$ .

A1A3L1L2

1.2

$a \ne b$ : the classes $\{(0,0)\}$ and $\{(0,1)\}$ are distinct members of the partition in [A2], and $q$ sends $(0,i)$ to the class of $(0,i)$ .

1.3

For $t \ne 0$ one has $q(t,0) = q(t,1)$ , the two points lying in the common class $\{(t,0),(t,1)\}$ .

2.1

Suppose $U, V \subseteq L$ are open with $a \in U$ , $b \in V$ and $U \cap V = \varnothing$ . Then $q^{-1}[U]$ and $q^{-1}[V]$ are open in $S$ by [A2], with $(0,0) \in q^{-1}[U]$ and $(0,1) \in q^{-1}[V]$ .

step 1.2A2assume-hyp

3.1

By [A1] the trace of $q^{-1}[U]$ at index $0$ is an open subset of $\mathbb{R}$ containing $0$ , so by [L1] there is $\varepsilon > 0$ with $(-\varepsilon,\varepsilon) \times \{0\} \subseteq q^{-1}[U]$ ; likewise there is $\delta > 0$ with $(-\delta,\delta) \times \{1\} \subseteq q^{-1}[V]$ .

step 2.1A1L1

4.1

Put $t := \min\{\varepsilon,\delta\}/2$ . Then $0 < t < \varepsilon$ and $t < \delta$ by [L1] and [L2], so $t \ne 0$ , $(t,0) \in q^{-1}[U]$ and $(t,1) \in q^{-1}[V]$ .

step 3.1L1L2

5.1

By step 1.3 and step 4.1 the point $q(t,0) = q(t,1)$ lies in $U$ and in $V$ , contradicting $U \cap V = \varnothing$ . So no such $U$ and $V$ exist.

step 1.3step 2.1step 4.1

6.1

By step 1.1 the space $S$ is Hausdorff, by [A2] the map $q$ is a quotient map, and by steps 1.2 and 5.1 the two distinct points $a$ and $b$ of $L$ have no disjoint open neighbourhoods, so $L$ is not Hausdorff by [A3]. The claim is therefore false.

step 1.1step 1.2step 5.1A2A3∎

Remarks

The source is not merely Hausdorff but metrizable, so strengthening the separation and countability properties of the source is not by itself what rescues the claim; what decides the matter is the relation being collapsed. A metric inducing the topology of $S$ is exhibited on the companion page, where the same witness is worked as 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 ↗, and the quotient map there is shown to be open as well.
The identification is as mild as it can be. Exactly one pair of points is left unidentified, and every other pair is glued; the failure is caused by two points that are not identified and yet have no disjoint saturated open neighbourhoods, since every neighbourhood of either origin contains a punctured interval that the other's neighbourhoods also contain.
What does survive is one direction of separation for the source. Nothing above says that a quotient of a Hausdorff space is badly behaved in general, and nothing here asserts which extra hypothesis on $q$ or on the relation restores the Hausdorff condition; that question belongs with the separation axioms, which are not available at this point in the reading order (What the theory of these constructions still owes at this point in the reading order: preservation of quotient maps under products, separation beyond Hausdorff, and the invariants that tell the glued spaces apart).

Depends on

Used by

Two copies of ℝ glued along ℝ ∖ {0} give a non-Hausdorff quotient of a metrizable space, by an open quotient map Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 84 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Hausdorff space (Wikipedia) (standard reference, not scraped)
Line with two origins (Wikipedia) (standard reference, not scraped)
Quotient space (topology) (Wikipedia) (standard reference, not scraped)