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In the subspace of $\mathbb{R}^2$ made of the vertical unit segments over $1/(n+1)$ together with the two points $(0,0)$ and $(0,1)$ , the component of $(0,0)$ is a singleton while its quasicomponent is $\{(0,0), (0,1)\}$

Statement refuted

Refuted: that the component and the quasicomponent of a point always agree. Every quasicomponent is a closed union of components, so each component is contained in a quasicomponent, and the quasicomponents partition the space proves only the inclusion $C(x) \subseteq Q(x)$ and asserts no converse; the inclusion can be strict.

Witness. In $\mathbb{R}^2$ with the product topology (For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$ , and hence also of $d_1$ and $d_2$ , so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space, The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space), and writing $\iota$ for the canonical natural so that $1/(n+1)$ means $1/\iota(n+1)$ with $n \in \mathbb{N}$ and $\mathbb{N}$ containing $0$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field), put

$S_n := \Bigl\{\tfrac{1}{n+1}\Bigr\} \times [0,1] \quad (n \in \mathbb{N}), \qquad a := (0,0), \qquad b := (0,1),$

$X \;:=\; \{a, b\} \;\cup\; \bigcup_{n \in \mathbb{N}} S_n ,$

with the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). Then

$C(a) = \{a\} \qquad \text{and} \qquad Q(a) = \{a, b\}$

(Connected components, quasicomponents, and totally disconnected spaces), so the inclusion is strict.

Facts & Assumptions

Given: $\mathbb{R}^2$ with the product topology and the subspace $X$ above, with $\pi_0, \pi_1$ the two projections.

[A1]

The sets $(c,d) \times (e,g)$ form a basis of $\mathbb{R}^2$ , and the open sets of $X$ are their traces together with unions of those; a map into $\mathbb{R}^2$ is continuous exactly when both components are; the projections are continuous (For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$ , and hence also of $d_1$ and $d_2$ , so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space, The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, Continuity of a map of topological spaces at a point and globally).

[A2]

A subset of $\mathbb{R}$ is a connected subset exactly when it is order-convex, and a continuous image of a connected space is a connected subset of the target (The connected subspaces of $\mathbb{R}$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $\mathbb{R}$ ", Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, A continuous image of a connected space is connected, and connectedness is a topological property, Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).

[A3]

$C(x)$ is the largest connected subset containing $x$ ; $Q(x)$ is the intersection of all clopen subsets containing $x$ ; $C(x) \subseteq Q(x)$ (Connected components, quasicomponents, and totally disconnected spaces, Every quasicomponent is a closed union of components, so each component is contained in a quasicomponent, and the quasicomponents partition the space, The components of a space are its maximal connected subsets, they partition it, and each of them is closed).

[A4]

A connected subset meeting a clopen set is contained in it, since the trace is clopen in it and nonempty (For a topological space the following agree: no separation exists, the only clopen subsets are $\varnothing$ and $X$ , and every continuous map to the two-point discrete space is constant, claim 2, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[A5]

$x \in \overline{A}$ exactly when every basic open set containing $x$ meets $A$ ; a set closed in $X$ contains its closure taken in $X$ , which is $\overline{A} \cap X$ (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set, Interior, closure, boundary, exterior, derived set and isolated point in a topological space, For $A \subseteq S \subseteq X$ the closure of $A$ in $S$ is $\overline{A}^{X} \cap S$ , while the interior only contains $\operatorname{int}^{X}(A) \cap S$ , with equality when $S$ is open; and a dense subset of $X$ traces to a dense subset of every open $S$ ).

[A6]

For every real $\varepsilon > 0$ there is a natural $k \ge 1$ with $1/k < \varepsilon$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

Counterexample

technique · direct

1.1

Each $S_n$ is a connected subset of $X$ : it is the image of $[0,1]$ , connected by [A2], under $t \mapsto (1/(n+1), t)$ , whose components are a constant map and the identity, both continuous by [A1].

A1A2

1.2

Write $P := \pi_0[X] = \{0\} \cup \{\, 1/(k+1) : k \in \mathbb{N} \,\}$ and, for $k \in \mathbb{N}$ , $\mu_k := \tfrac12\bigl(1/(k+2) + 1/(k+1)\bigr)$ , which satisfies $1/(k+2) < \mu_k < 1/(k+1)$ and lies in no $P$ , being strictly between two consecutive members of $P$ and strictly positive. No subset of $P$ with two distinct points is order-convex: given $p < q$ in $P$ , if $p = 0$ and $q = 1/(m+1)$ take $k := m$ , and if $p = 1/(n+1)$ and $q = 1/(m+1)$ with $m < n$ take $k := n-1$ , which is a natural number since $n \ge 1$ ; in both cases $p \le 1/(k+2) < \mu_k < 1/(k+1) \le q$ , so $\mu_k$ lies strictly between $p$ and $q$ and outside $P$ .

A2A6

1.3

Each $S_n$ is clopen in $X$ : with $\delta := \tfrac12\bigl(\tfrac{1}{n+1} - \tfrac{1}{n+2}\bigr) > 0$ , the trace on $X$ of the open strip $(1/(n+1) - \delta,\ 1/(n+1) + \delta) \times \mathbb{R}$ is exactly $S_n$ , so $S_n$ is open in $X$ ; and the trace of the closed strip $[1/(n+1) - \delta,\ 1/(n+1)+\delta] \times \mathbb{R}$ is also exactly $S_n$ , so $S_n$ is closed in $X$ by [A5].

A1A5

1.4

$b \in Q(a)$ . Let $K$ be clopen in $X$ with $a \in K$ . Since $K$ is open, [A1] gives $\varepsilon > 0$ with $X \cap \bigl((-\varepsilon,\varepsilon) \times (-\varepsilon,\varepsilon)\bigr) \subseteq K$ ; by [A6] there is $k \ge 1$ with $1/k < \varepsilon$ , and putting $n := k-1 \in \mathbb{N}$ the point $(1/(n+1), 0)$ lies in that trace, hence in $K$ . The same holds for every $m \ge n$ , since $1/(m+1) \le 1/(n+1) < \varepsilon$ .

A1A6

2.1

$C(a) = \{a\}$ . Let $E \subseteq X$ be connected with $a \in E$ . Then $\pi_0[E]$ is a connected subset of $\mathbb{R}$ by [A1] and [A2], hence order-convex, and it lies inside $\pi_0[X]$ ; by step 1.2 it has at most one point, so $\pi_0[E] = \{0\}$ and $E \subseteq \{a,b\}$ . But $\{a\}$ is the trace on $\{a,b\}$ of the open set $\mathbb{R} \times (-1/2, 1/2)$ , so $\{a\}$ and $\{b\}$ are both open in the subspace $\{a,b\}$ ; a connected $E$ containing $a$ therefore cannot contain $b$ , and $E = \{a\}$ . Hence $C(a) = \{a\}$ by [A3].

step 1.2A1A2A3

2.2

$Q(a) \subseteq \{a,b\}$ : for each $n$ the set $X \setminus S_n$ is clopen by step 1.3 and contains $a$ , so $Q(a) \subseteq X \setminus S_n$ by [A3]; intersecting over all $n$ leaves $Q(a) \subseteq X \setminus \bigcup_n S_n = \{a,b\}$ .

step 1.3A3

2.3

Each such $S_m$ is connected by step 1.1 and meets $K$ at $(1/(m+1), 0)$ , so $S_m \subseteq K$ by [A4]; in particular $(1/(m+1), 1) \in K$ for every $m \ge n$ .

step 1.1step 1.4A4

3.1

Every basic open set containing $b = (0,1)$ contains, for large enough $m$ , the point $(1/(m+1), 1)$ : such a set includes $(-\eta,\eta) \times (1-\eta, 1+\eta)$ for some $\eta > 0$ , and [A6] supplies $m \ge n$ with $1/(m+1) < \eta$ . So $b \in \overline{K}$ by [A5], and $K$ being closed in $X$ gives $b \in K$ . As $K$ was an arbitrary clopen set containing $a$ , this shows $b \in Q(a)$ .

step 2.3A5A6

4.1

With step 2.2 and $a \in Q(a)$ from [A3], $Q(a) = \{a,b\}$ , while $C(a) = \{a\}$ by step 2.1. So $C(a) \subsetneq Q(a)$ and the two notions differ.

step 2.1step 2.2step 3.1A3∎

Remarks

Why no clopen set can separate $a$ from $b$ . A clopen set containing $a$ must, by openness, catch a point of $S_n$ for every large $n$ ; being clopen it must then swallow each of those whole segments, by [A4]; and being closed it must contain the limit of their top endpoints, which is $b$ . The segments act as a ladder that is invisible to connectedness — no connected set climbs it, since $\pi_0$ would have to be order-convex — and unavoidable for clopen sets.
The two points are essential and so are the segments. Removing $b$ makes $Q(a) = \{a\}$ ; replacing the segments by single points makes $X$ totally separated, because each column becomes clopen on its own and no ladder survives. This is why the witness needs sets that are connected and shrinking towards the limit, not merely a sequence of points.
The space has no isolated ladder rung near the limit. By step 1.3 each $S_n$ is clopen, so $X$ is not connected; the failure recorded here is not about $X$ being connected but about the two ways of measuring how $X$ falls apart giving different answers at $a$ .

Depends on

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Sources

Locally connected space (Wikipedia) (standard reference, not scraped)
Connected space (Wikipedia) (standard reference, not scraped)
The Stacks Project, Section 5.7: Connected components (standard reference, not scraped)

In the subspace of R2\mathbb{R}^2R2 made of the vertical unit segments over 1/(n+1)1/(n+1)1/(n+1) together with the two points (0,0)(0,0)(0,0) and (0,1)(0,1)(0,1), the component of (0,0)(0,0)(0,0) is a singleton while its quasicomponent is {(0,0),(0,1)}\{(0,0), (0,1)\}{(0,0),(0,1)}