Connectedness: Examples and Counterexamples

Let $n\in \mathbb{N}$ with $n\ge 1$ and give ${\mathbb{R}}^n$ the product topology, which is the metric topology of $d_{\mathrm{\infty }}$ (For $n\ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_{\mathrm{\infty }}$ 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, ${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$, and $d_1$, $d_2$, $d_{\mathrm{\infty }}$ are metrics on it, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement). Recall that ${\mathbb{R}}^n$ is a real vector space under coordinatewise operations (Vector space over a field, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

A subset $C\subseteq {\mathbb{R}}^n$ is convex when

$x,y\in C\text{ and }t\in [0,1]⟹(1-t)x+ty\in C$

(Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). Then:

1. Every convex $C\subseteq {\mathbb{R}}^n$ is path-connected (Paths, path-connected spaces and path components), hence connected (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets, Every path-connected space is connected, and every path component lies inside a component).
2. Every ball is convex, in each of the norms $\parallel \cdot {\parallel }_1$, $\parallel \cdot {\parallel }_2$, $\parallel \cdot {\parallel }_{\mathrm{\infty }}$ (The $p$-norms $\parallel x{\parallel }_p$ for rational $p\ge 1$, and $\parallel x{\parallel }_{\mathrm{\infty }}$, Open ball, closed ball and sphere in a metric space); so every ball of ${\mathbb{R}}^n$ is path-connected and connected.
3. ${\mathbb{R}}^n$ itself is convex, hence path-connected and connected, and so is every half-space $\{x:x_k\le c\}$, and every box ${\prod }_{k<n}{J}_k$ with each $J_k$ an order-convex subset of $\mathbb{R}$.

• The path is the straight segment and nothing more is needed. Convexity is exactly the hypothesis that the segment between two points of the set stays in the set, so the definition of the path writes itself; the only work is that the segment is a continuous map, which is [A1] plus the continuity of an affine map of one real variable.

• Convexity is far from necessary. A circle is path-connected and not convex, and so is any set obtained from a convex one by bending it. Nothing above asserts a converse.

• The hypothesis $n\ge 1$ comes from $d_{\mathrm{\infty }}$. $\parallel \cdot {\parallel }_{\mathrm{\infty }}$ is a maximum over $n$ terms and is undefined at $n=0$ (The $p$-norms $\parallel x{\parallel }_p$ for rational $p\ge 1$, and $\parallel x{\parallel }_{\mathrm{\infty }}$, ${\mathbb{R}}^n$ as the set of functions $n\to \mathbb{R}$, and $d_1$, $d_2$, $d_{\mathrm{\infty }}$ are metrics on it). At $n=0$ the product is a one-point space, which is path-connected outright.

Let $X$ be a set carrying one of the six topologies of The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies. The table records where each sits, with connectedness as in Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets, local connectedness as in Locally connected and locally path-connected spaces: a neighbourhood base of open connected, respectively open path-connected, sets at every point and path-connectedness as in Paths, path-connected spaces and path components.

Sierpinski space is the particular-point topology on a two-point set (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), so its row is an instance of the row above it and is not verified separately.

The component column reads at nonempty $X$; the empty space is connected and has no components at all, there being no points (The components of a space are its maximal connected subsets, they partition it, and each of them is closed).

Two entries are deliberately left open. No item among this page's declared prerequisites settles whether the cofinite topology on an infinite $X$, or the cocountable topology on an uncountable $X$, is path-connected, so the table says nothing either way.

• Connectedness is cheap when there are few open sets. Four of the six topologies are connected for the same structural reason: no two nonempty open sets are disjoint. That is immediate for the indiscrete and particular-point topologies, and for the cofinite and cocountable ones it is the statement that the ambient set is not a union of two small sets.

• Local connectedness here is never informative. In every connected case above, every nonempty open subspace is again of the same kind and hence connected, so local connectedness holds for free. A space where local connectedness carries information must have open sets that are not themselves connected, which is what happens in ${\mathbb{R}}^2$ and its subspaces.

• Why two cells are left blank. A path in the cofinite or cocountable topology is a continuous map out of $[0,1]$, and deciding whether a non-constant one exists needs machinery this page's declared prerequisites do not supply: for the cofinite case a comparison of $\mathrm{\mid }X\mathrm{\mid }$ with the cardinality of $\mathbb{R}$, and for the cocountable case an argument about the image of a dense countable subset. Neither is available here, so the honest entry is that the question is not settled.

Let $G$ be the graph of the zigzag function and $\bar{G}=G\cup (\{0\}\times [0,1])$ its closure in ${\mathbb{R}}^2$ (The graph of the piecewise-linear map oscillating between $0$ and $1$ on the intervals $[1\mathrm{/}(n+2),1\mathrm{/}(n+1)]$ is path-connected, its closure adds the segment $\{0\}\times [0,1]$, and that closure is connected, is not path-connected because no path joins the segment to the graph, and is not locally connected, claim 2). Write $\mathrm{\Sigma }:=\{0\}\times [0,1]$ for the added segment. Then, in the space $\bar{G}$ with its 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):

1. One component. $\bar{G}$ is connected, so it has exactly one component, namely $\bar{G}$ itself, and exactly one quasicomponent, also $\bar{G}$ (Connected components, quasicomponents, and totally disconnected spaces, The components of a space are its maximal connected subsets, they partition it, and each of them is closed, Every quasicomponent is a closed union of components, so each component is contained in a quasicomponent, and the quasicomponents partition the space).
2. Two path components, namely $G$ and $\mathrm{\Sigma }$ (Paths, path-connected spaces and path components).
3. $G$ is open in $\bar{G}$ and $\mathrm{\Sigma }$ is closed in it; neither is clopen, since $\bar{G}$ is connected.
4. Local connectedness holds at every point of $G$ and fails at every point of $\mathrm{\Sigma }$ (Locally connected and locally path-connected spaces: a neighbourhood base of open connected, respectively open path-connected, sets at every point). So the set of points at which $\bar{G}$ fails to be locally connected is exactly $\mathrm{\Sigma }$.

Claim 2 is the sharp form of "not path-connected": the failure is not that some pair of points is unjoined but that the space splits into exactly two path classes, one of which is the whole added segment.

• The two path components have different topological characters. $\mathrm{\Sigma }$ is homeomorphic to a closed bounded interval and $G$ to a half-open one, and only the first is closed in $\bar{G}$. So the partition into path components is not a partition into clopen pieces, which is exactly what A connected, locally path-connected space is path-connected, because its path components are open would supply if the space were locally path-connected — and it is not, by claim 4.

• Why $\bar{G}$ has one component and two path components at once. Connectedness is destroyed only by a clopen splitting, and adjoining $\mathrm{\Sigma }$ to $G$ creates none: every neighbourhood of a point of $\mathrm{\Sigma }$ meets $G$ by 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, claim 1, and claim 2 of The graph of the piecewise-linear map oscillating between $0$ and $1$ on the intervals $[1\mathrm{/}(n+2),1\mathrm{/}(n+1)]$ is path-connected, its closure adds the segment $\{0\}\times [0,1]$, and that closure is connected, is not path-connected because no path joins the segment to the graph, and is not locally connected. Path-connectedness is destroyed by a single continuity failure along one map, and that is enough.

• The quasicomponent carries no extra information here. Since the space is connected, its unique quasicomponent equals its unique component, so this witness says nothing about the strictness of the inclusion $C(x)\subseteq Q(x)$; a separate space is needed for that.

Let $\mathbb{Q}={\mathbb{Q}}_{\mathbb{R}}$ be the copy of the rationals inside $\mathbb{R}$ (Both $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable) with the subspace topology of the usual 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, 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). Then:

1. Every component is a single point: $C(q)=\{q\}$ for every $q\in \mathbb{Q}$, so $\mathbb{Q}$ is totally disconnected (Connected components, quasicomponents, and totally disconnected spaces).
2. No point is isolated: $\{q\}$ is not open in $\mathbb{Q}$, so the topology is not discrete (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).
3. $\mathbb{Q}$ is not locally connected at any of its points.
4. The components are closed and not open. They are closed by The components of a space are its maximal connected subsets, they partition it, and each of them is closed and not open by claim 2, so $\mathbb{Q}$ is a space in which every component fails to be clopen.

$\mathbb{Q}$ is countably infinite ($\mathbb{Q}$ is countably infinite) while its complement in $\mathbb{R}$ is uncountable (The irrationals are uncountable); it is the abundance of the complement, not the scarcity of $\mathbb{Q}$, that drives claim 1.

• Total disconnectedness and discreteness come apart here, and the two steps that separate them are step 2.1 and step 2.2. The first says the irrationals block every interval, so no connected set can span two rationals; the second says the rationals themselves are everywhere, so no rational is isolated. A space can be shredded into points without those points being separated.

• Local connectedness fails for a structural reason, not a delicate one. In any totally disconnected space the only candidates for a connected neighbourhood are singletons, so local connectedness at $x$ is equivalent to $\{x\}$ being open. Hence a totally disconnected space is locally connected exactly when it is discrete, and step 3.1 is that observation applied to $\mathbb{Q}$.

• The same argument applies to the irrationals. Nothing in steps 1.1 and 2.1 used countability of $\mathbb{Q}$; only that its complement meets every interval. The irrationals have that property too by [A3], so they are totally disconnected and not discrete as well, and they are uncountable (The irrationals are uncountable).

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 $d_{\mathrm{\infty }}$ 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 $1\mathrm{/}(n+1)$ means $1\mathrm{/}\iota (n+1)$ with $n\in \mathbb{N}$ and $\mathbb{N}$ containing $0$ (The canonical natural $\iota (n)=n\cdot 1_F$ of a field), put

$S_n:=\left\{\frac{1}{n+1}\right\}\times [0,1](n\in \mathbb{N}),a:=(0,0),b:=(0,1),$

$X:=\{a,b\}\cup {\bigcup }_{n\in \mathbb{N}}{S}_n,$

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

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

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

• 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 $a$.

Let $R={\omega }_1\times [0,1)$ be the closed long ray with its lexicographic order and its order topology (The closed long ray ${\omega }_1\times [0,1)$ under the lexicographic order, and the long line, with the order topology, The order topology of a linearly ordered set, with the open rays as a subbasis; order-convex sets, order-density, the least upper bound property, and linear continua), and let $0_R=(0,0)$ be its least element. For $x\in R$ write $[0_R,x]:=\{y\in R:y\le x\}$ for the initial segment up to $x$. Then:

1. $R$ is connected, and its unique component is $R$ (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets, Connected components, quasicomponents, and totally disconnected spaces).
2. Every initial segment $[0_R,x]$ is order-convex and connected, and so is every open ray and every interval of $R$.
3. $R$ is locally connected (Locally connected and locally path-connected spaces: a neighbourhood base of open connected, respectively open path-connected, sets at every point).
4. Assuming the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)), no at most countable subset of $R$ is cofinal in $R$, that is, every at most countable subset has a strict upper bound (Finite, countably infinite, countable, uncountable).

Claim 4 is the order-theoretic analogue, transported to $R$, of the statement that no at most countable subset of ${\omega }_1$ is cofinal in ${\omega }_1$ (Assuming countable choice: every at most countable subset of ${\omega }_1$ is bounded below ${\omega }_1$, so no at most countable subset of ${\omega }_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable, Cofinal subset of an ordinal); here a subset $D\subseteq R$ is called cofinal when for every $x\in R$ there is $y\in D$ with $x\le y$.

Path-connectedness is not asserted. Whether $R$ is path-connected is not settled by any item among this page's declared prerequisites, and nothing here claims it either way. Consequently the path components of $R$ are not computed.

• Local connectedness is immediate here and is not a deep property of $R$. Every basic open set of an order topology is order-convex, and in a linear continuum every order-convex set is connected. So any linear continuum is locally connected, and $R$ inherits that with no reference to ${\omega }_1$.

• What distinguishes $R$ from an ordinary half-line is claim 4 alone. The first three claims hold verbatim for $[0,\mathrm{\infty })\subseteq \mathbb{R}$, which is also a linear continuum with a least element and no greatest. In $[0,\mathrm{\infty })$ the at most countable set of canonical naturals is cofinal; in $R$ no at most countable set is, and that is the whole content of the word long.

• The choice cost is inherited and is not spent again here. Claim 4 uses claim 3 of The long ray is a linear continuum, hence connected; every one of its at most countable subsets is bounded above, assuming countable choice, whose own statement carries ${\mathrm{A}\mathrm{C}}_{\omega }$; the argument above adds only the passage from an upper bound to a strict one, which needs nothing beyond $R$ having no greatest element.