nLab connected space

Idea

A topological space is connected if it can not be split up into two independent parts.

Every topological space may be decomposed into disjoint maximal connected subspaces, called its connected components. The underlying set of a topological space is the disjoint union of the underlying sets of its connected components, but the space itself is not necessarily the coproduct of its connected components in the category Top of topological spaces.

One often studies topological ideas first for connected spaces and then generalises to general spaces. This is especially true if one is studying such nice topological spaces that every space is a coproduct of connected components (such as for example locally connected topological spaces).

Definitions

Classical definition

Definition

(connected topological space)

A topological space $(X, \tau)$ is connected if the following equivalent conditions hold:

1. For all pairs of topological spaces $(X_1, \tau_1), (X_2, \tau_2)$ such that $(X, \tau)$ is homeomorphic to their disjoint union space

   $$(X,\tau) \,\simeq\, (X_1,\tau_1) \sqcup (X_2,\tau_2) \,,$$

   exactly one of the two spaces is the empty space.

2. For all pairs of open subsets $U_1, U_2 \subset X$, if

   $$U_1 \cup U_2 \,=\, X \phantom{A}\text{and} \phantom{A} U_1 \cap U_2 \,=\, \emptyset$$

   then exactly one of the two subsets is the empty set.

3. If a subset $CO \subseteq X$ is clopen (both closed and open), then $CO = X$ if and only if $CO$ is inhabited.

Remark

According to def. the empty topological space is not regarded as connected. Some authors do want the empty space to count as a connected space. This means to change in the two first items of def. the “exactly one” to “at least one” and in the third item “if and only if” to “if”.

Proposition

The conditions in def. are indeed equivalent.

Proof

First consider the equivalence of the first two statements:

Suppose that in every disjoint union decomposition of $(X,\tau)$ exactly one summand is empty. Now consider two disjoint open subsets $U_1, U_2 \subset X$ whose union is $X$ and whose intersection is empty. We need to show that exactly one of the two subsets is empty.

Write $(U_1, \tau_{1})$ and $(U_2, \tau_2)$ for the corresponding topological subspaces. Then observe that from the definition of subspace topology and the disjoint union space we have a homeomorphism

$$X \simeq (U_1, \tau_1) \sqcup (U_2, \tau_2)$$

because by assumption every open subset $U \subset X$ is the disjoint union of open subsets of $U_1$ and $U_2$, respectively:

$$U = U \cap X = U \cap (U_1 \sqcup U_2) = (U \cap U_1) \sqcup (U \cap U_2) \,,$$

which is the definition of the disjoint union topology.

Hence by assumption exactly one of the two summand spaces is the empty space and hence the underlying set is the empty set.

Conversely, suppose that for every pair of open subsets $U_1, U_2 \subset U$ with $U_1 \cup U_2 = X$ and $U_1 \cap U_2 = \emptyset$ then exactly one of the two is empty. Now consider a homeomorphism of the form $(X,\tau) \simeq (X_1, \tau_1) \sqcup (X_2,\tau_2)$. By the nature of the disjoint union space this means that $X_1, X_2 \subset X$ are disjoint open subsets of $X$ which cover $X$. So by asumption precisely one of the two subsets is the empty set and hence precisely one of the two topological spaces is the empty space.

Now regarding the equivalence to the third statement:

If a subset $CO \subset X$ is both closed and open, this means equivalently that it is open and that its complement $X \setminus CO$ is also open, hence equivalently that there are two open subsets $CO, X \setminus CO \subset X$ whose union is $X$ and whose intersection is empty. This way the third condition is equivalent to the second.

Constructive definition

In constructive mathematics the conditions in Def. no longer need to be equivalent. See Taylor 2010, Def. 13.2, who essentially speaks of “compact connectedness” for the first item in Def. and of “overt connectedness” for the second.

Related discussion in terms of cohesive homotopy type theory (and for the special case of the Dedekind real numbers) is in Shulman 2018, Thm. 11.1, 11.3.

In this language of homotopy type theory one might then say:

A 0-truncated cohesive infinity-groupoid is “compact connected” (in the sense of Taylor 2010, Def. 13.2), if for subspaces $A \subseteq S$ and $B \subseteq S$ of $S$ with embeddings $i_{A,S}:A \hookrightarrow S$ and $i_{B,S}:B \hookrightarrow S$ such that the embedding $i_{A \cup B,S}:A \cup B \hookrightarrow S$ is an equivalence $(i_{A \cup B,S}:A \cup B \simeq S)$ and the embedding $i_{\emptyset,A \cap B}:\emptyset \hookrightarrow A \cap B$ is an equivalence $(i_{\emptyset,A \cap B}:\emptyset \simeq A \cap B)$, either $i_{A,S}$ is an equivalence ($i_{A,S}:A \simeq S$) or $i_{B,S}$ is an equivalence ($i_{B,S}:B \simeq S$).

Category-theoretic definition

In the language of category theory def. may be rephrased as follows:

Let Top denote the category of all topological space.

Then a topological space $X$ is connected precisely if the representable functor

$$hom(X, -) \;\colon\; Top \longrightarrow Set$$

preserves all coproducts.

This, in turn, is equivalent to just requiring that the functor preserves binary coproducts (a detailed proof of this, in a more general setting, is given at connected object in this proposition. In that case, notice that we always have a map

$$hom(X,Y) + hom(X,Z) \longrightarrow hom(X,Y + Z) ,\,$$

so $X$ is connected if this is always a bijection. This definition generalises to the notion of connected objects in an extensive category.

The variant of the definition according to remark , with regards to the empty space, means, in terms of category theoretic language, that one requires only that the maps

$$hom(X,Y) + hom(X,Z) \to hom(X,Y + Z)$$

be surjections.

However, many results come out more cleanly by disqualifying the empty space as connected (much as one disqualifies $1$ when one defines the notion of prime number). See also the discussion at empty space and at too simple to be simple.

Connected components

Every topological space $X$ admits an equivalence relation $\sim$ where $x \sim y$ means that $x$ and $y$ belong to some subspace which is connected. The equivalence class $Conn(x)$ of an element $x$ is thus the union of all connected subspaces containing $x$; it follows readily from Result that $Conn(x)$ is itself connected. It is called the connected component of $x$. It is closed, by Result . A space is connected if and only if it has exactly one connected component (or at most one, if you allow the empty space to be connected).

Definition

(connected components)

For $(X,\tau)$ a topological space, its connected components are the equivalence classes under the equivalence relation on $X$ which regards two points as equivalent if they both sit in some open subset which, as a topological subspace (example ), is connected (def. ):

$$(x \sim y) \;\coloneqq\; \left( \underset{ U \subset X \; \text{\open} }{\exists} \left( \left( x,y \in U \right) \phantom{A}\text{and}\phantom{A} \left( U \, \text{is connected} \right) \right) \right) \,.$$

Remark

(quasi-components)

There is another equivalence relation $\sim_q$ modelling a notion of connected components, where $x \sim_q y$ iff $f(x) = f(y)$ for every continuous map $f \colon X \to D$ to a discrete space $D$.

The corresponding equivalence class of $x$ may alternatively be described as the intersection of all clopen subsets that contain $x$. This is called the quasi-component of $x$, denoted here as $QConn(x)$.

It is easy to prove that this relates to the equivalence classes $Conn(x)$ from Def. as

$$Conn(x) \subseteq QConn(x)$$

and that equality holds if $X$ is compact Hausdorff or is locally connected, but also in other circumstances (such as the space of rational numbers as a subspace of the real line).

Example

For an example where $Conn(x) \neq QConn(x)$, take $X$ to be the following subspace of $[0, 1] \times [0, 1]$:

$$X = \{(0, 0), (0, 1)\} \cup \bigcup_{n \geq 1} \{1/n\} \times [0, 1]$$

In this example, $Conn((0, 1)) = \{(0, 1)\}$, but $QConn((0, 1)) = \{(0, 0), (0, 1)\}$.

Remark

Warning

It is not generally true that a topological space is the disjoint union space (coproduct in Top) of its connected components, nor of its quasi-components.

The spaces such that all their open subspaces are the disjoint union of their connected components are the locally connected topological spaces.

Example

The connected components in Cantor space $2^{\mathbb{N}}$ (with its topology as a product of 2-point discrete spaces) are just the singletons, but the coproduct of the singleton subspaces carries the discrete topology, which differs from that of Cantor space.

Similarly for set of rational numbers with its absolute-value topology (the one induced as a topological subspace of the real line).

Examples

Basic examples

Example

(continuous images of connected spaces are connected)

The regular image of a connected space $X$ under a continuous map $f: X \to Y$ (i.e., the set-theoretic image with the subspace topology inherited from $Y$) is connected. Or, what is essentially the same: if $X$ is connected and $f: X \to Y$ is epic, then $Y$ is connected:

Let $X$ be a connected topological space, let $Y$ be any topological space, and let

$$f \;\colon\; X \longrightarrow Y$$

be a continuous function. This factors via continuous functions through the image

$$f \;\colon\; X \underoverset{surjective}{p}{\longrightarrow} f(X) \underoverset{injective}{i}{\longrightarrow} Y$$

for $f(X)$ equipped either with the subspace topology relative to $Y$ or the quotient topology relative to $X$. In either case:

If $X$ is a connected topological space, then so is $f(X)$.

Proof

Let $U_1,U_2 \subset f(X)$ be two open subsets such that $U_1 \cup U_2 = f(X)$ and $U_1 \cap U_2 = \emptyset$. We need to show that precisely one of them is the empty set.

Since $p$ is a continuous function, also the pre-images $p^{-1}(U_1), p^{-1}(U_2) \subset X$ are open subsets and are still disjoint. Since preimages preserve unions it also follows that $p^{-1}(U_1) \cup p^{-1}(U_2) = X$. Since $X$ is connected, it follows that one of these two pre-images $p^{-1}(U_i)$ is the empty set. Since $p$ is surjective, this implies that $U_i$ is empty, which means that $f(X)$ is connected.

Example

Wide pushouts of connected spaces are connected. (This would of course be false if the empty space were considered to be connected.) This follows from the hom-functor definition of connectedness, plus the fact that coproducts in $Set$ commute with wide pullbacks. More memorably: connected colimits of connected spaces are connected.

Example

If $S \subseteq X$ is a connected subspace and $S \subseteq T \subseteq \overline{S}$ (i.e. if $T$ is between $S$ and its closure), then $T$ is connected. Or, what is essentially the same: if $T$ has a dense connected subspace $S$, then $T$ is connected.

(see also prop. below)

Example

(product space of connected spaces is connected)

Let $\{X_i\}_{i \in I}$ be a set of connected spaces. Then also their product topological space $\underset{i \in I}{\prod}X_i$ (with the Tychonoff topology) is connected.

Proof

This relies on some special features of Top. A general abstract proof is given at connected object in this theorem and this remark.

Here is an alternative elementary proof in point-set topology:

Let $U_1, U_2 \subset \underset{i \in I}{\prod}X_i$ be an open cover of the product space by two disjoint open subsets. We need to show that precisely one of the two is empty. Since each $X_i$ is connected and hence non-empty, the product space is not empty, and hence it is sufficient to show that at lest one of the two is empty.

Assume on the contrary that both $U_1$ and $U_2$ are non-empty.

Observe first that if so, then we could find $x_1 \in U_1$ and $x_2 \in U_2$ whose coordinates differed only a a finite subset of $I$. This is since by the nature of the Tychonoff topology $\pi_i(U_1) = X_i$ and $\pi_i(U_2) = X_i$ for all but a finite number of $i \in iI$.

Next observe that we then could even find $x'_1 \in U_1$ that differed only in a single coordinate from $x_2$: Because pick one coordinate in which $x_1$ differs from $x_2$ and change it to the corresponding coordinate of $x_2$. Since $U_1$ and $U_2$ are a cover, the resulting point is either in $U_1$ or in $U_2$. If it is in $U_2$, then $x_1$ already differed in only one coordinate from $x_2$ and we may take $x'_1 \coloneqq x_1$. If instead the new point is in $U_1$, then rename it to $x_1$ and repeat the argument. By induction this finally yields an $x'_1$ as claimed.

Therefore it is now sufficient to see that it leads to a contradiction to assume that there are points $x_1 \in U_1$ and $x_2 \in U_2$ that differ in only the $i_0$th coordinate, for some $i_0 \in I$ then $x_1 = x_2$.

Observe that the inclusion

$$\iota \colon X_{i_0} \longrightarrow \underset{i \in I}{\prod} X_i$$

which is the identity on the $i_0$th component and is otherwise constant on the $i$th component of $x_1$ or equivalently of $x_2$ is a continuous function, by the nature of the Tychonoff topology.

Therefore also the restrictions $\iota^{-1}(U_1)$ and $\iota^{-1}(U_2)$ are open subsets. Moreover they are still disjoint and cover $X_i$. Hence by the connectedness of $X_i$, precisely one of them is empty. This means that the $i_0$-component of both $x_1$ and $x_2$ must be in the other subset of $X_i$, and hence that $x_1$ and $x_2$ must both be in $U_1$ or both in $U_2$, contrary to the assumption.

Example

(connected subspaces of the real line are the intervals)

Regard the real line with its Euclidean metric topology. Then a subspace $S \subset \mathbb{R}$ is connected (def. ) precisely if it is an interval, hence precisely if

$$\underset{x,y \in S \subset \mathbb{R}}{\forall} \underset{ r \in \mathbb{R} }{\forall} \left( \left( x \lt r \lt y \right) \Rightarrow \left( r \in S \right) \right) \,.$$

In particular for $\{ I_i \subset \mathbb{R} \}_{i \in I}$ a set of disjoint intervals, then $I$ is the set of connected components of the union $\underset{i \in I}{\cup} I_i$.

Proof

Suppose on the contrary that we have $x \lt r \lt y$ but $r \notin S$. Then by the nature of the subspace topology there would be a decomposition of $S$ as a disjoint union of disjoint open subsets:

$$S = \left( S \cap (r,\infty) \right) \sqcup \left( S \cap (-\infty,r) \right) \,.$$

But since $x \lt r$ and $r \lt y$ both these open subsets were non-empty, thus contradicting the assumption that $S$ is connected. This yields a proof by contradiction.

Exotic examples

The basic results above give a plethora of ways to construct connected spaces. More exotic examples are sometimes useful, especially for constructing counterexamples.

Example

The following, due to Bing, is a countable connected Hausdorff space. Let $Q = \{(x, y) \in \mathbb{Q} \times \mathbb{Q}: y \geq 0\}$, topologized by defining a basis of neighborhoods $N_{\epsilon, a, b}$ for each point $(a, b) \in Q$ and $\epsilon \gt 0$:

$$N_{a, b} \coloneqq \{(a, b)\} \cup \{(s, 0) \in Q: {|a + b/\theta - s|} \lt \epsilon\} \cup \{(s, 0) \in Q: {|a - b/\theta - s|} \lt \epsilon\}$$

where $\theta \lt 0$ is some chosen fixed irrational number. It is easy to see this space is Hausdorff (using the fact that $\theta$ is irrational). However, the closure of $N_{\epsilon, a, b}$ consists of points $(x, y)$ of $Q \times Q$ with either $(x-a) - \epsilon \leq (y-b)/\theta \leq (x-a) + \epsilon$ or $(x-a) - \epsilon \leq -(y-b)/\theta \leq (x-a) + \epsilon$, in other words, the union of two infinitely long strips of width $2\epsilon$ and slopes $\theta$, $-\theta$. Clearly any two such closures intersect, and therefore the space is connected.

Example

This example is due to Golomb. Topologize the set of natural numbers $\mathbb{N}$ by taking a basis to consist of sets $A_{a,b} \coloneqq \{a k + b | k = 1,2, \ldots\}$, where $a, b \in \mathbb{N}$ are relatively prime. The space is Hausdorff, but the intersection of the closures of two non-empty open sets is never empty, so this space is connected.

Properties

Example

(locally constant function on connected topological spaces are constant functions)

If $X$ is a connected topological space and $f \colon X \to Y$ is a locally constant continuous function, then $f$ is in fact a constant function.

Proof

By definition of locally constant functions, every point $x \in X$ has an open neighborhood $U_x$ such that the restriction $f\vert_{U_x}$ is a constant function. The unions of these neighborhoods for a fixed constant value hence are disjoint open subsets that constitute a cover of $X$. By connectedness this cover must consist of a single non-empty element. But by construction this means that $f$ is constant.

Path-connectedness

An important variation on the theme of connectedness is path-connectedness.

Definition

(continuous path in a topological space)

Let $(X,\tau)$ be a topological space, Then a continuous path (or just path, for short) in $(X,\tau)$ is a continuous function of the form

$$\gamma \;\colon\; [0,1] \longrightarrow (X,\tau)$$

where the domain is the closed interval equipped with its Euclidean subspace topology.

One says that the path connects the point $\gamma(0) \in X$ with the point $\gamma(1) \in X$.

For $x \in X$ a fixed point, then the subset

$$PConn_x(X) \;\coloneqq\; \left\{ y \in X \;\vert\; \underset{\text{path}\, \gamma}{\exists} \left( \left( \gamma(0) = x \right) \phantom{} \text{and} \phantom{A} \left( \gamma(1) = y \right) \right) \right\}$$

is called the path-connected component of $x$.

The set of path connected components of $X$ is denoted

$$\pi_0(X) \,.$$

The set $\pi_0(X)$ of path components (the 0th “homotopy group”) is thus the coequalizer in

(1)$$\hom([0, 1], X) \underoverset {ev_1} {ev_0} {\rightrightarrows} \hom(1, X) \to \pi_0(X) .$$

Observe that this is a reflexive coequalizer, as witnessed by the mutual right inverse $\hom(!, X): \hom(1, X) \to \hom([0, 1], X)$.

(We can even topologize $\pi_0(X)$ by taking the coequalizer in $Top$ of

$$X^{[0, 1]} \stackrel{\overset{ev_0}{\to}}{\underset{ev_1}{\to}} X,$$

taking advantage of the fact that the locally compact Hausdorff space $[0, 1]$ is exponentiable. The resulting quotient space will be discrete if $X$ is locally path-connected.)

We say $X$ is path-connected if it has exactly one path component.

It follows easily from the basic results above that:

Lemma

A path connected space $X$ is connected.

Proof

Assume it were not, then it would be covered by two disjoint inhabited open subsets $U_1, U_2 \subset X$. But by path connectedness there were a continuous path $\gamma \colon [0,1] \to X$ from a point in one of the open subsets to a point in the other. The continuity would imply that $\gamma^{-1}(U_1), \gamma^{-1}(U_2) \subset [0,1]$ were a disjoint open cover of the interval. This would be in contradiction to the fact that intervals are connected. Hence we have a proof by contradiction.

However, it need not be closed (and therefore need not be the connected component of $x$); see the following example. The path components and connected components do coincide if $X$ is locally path-connected.

Example

The topologist’s sine curve

$$\{ (x, y) \in \mathbb{R}^2 \;:\; (0 \lt x \leq 1 \;\wedge\; y = sin(1/x)) \;\vee\; (0 = x \;\wedge\; -1 \leq y \leq 1) \}$$

provides a classic example where the path component of a point need not be closed. (Specifically, consider a point on the locus of $y = \sin(1/x)$.)

The basic categorical Results , , and above carry over upon replacing “connected” by “path-connected”. (As of course does example , trivially.)

Finally, as a contrast to a path-connected space, a totally path-disconnected space is a space such that its set of path components is equal to the underlying set of the space. Equivalently, that there are no non-constant paths. This by far does not mean that the space is discrete!

Arc-connectedness

A refinement of the notion of path-connected space is that of arc-connected (or arcwise-connected) space:

Definition

A space $X$ is arc-connected if for any two distinct $x, y \in X$ there exists an injective continuous map $\alpha: I \to X$ such that $\alpha(0) = x$ and $\alpha(1) = y$.

Arc-connected spaces are of course path-connected, but there are trivial examples (using an indiscrete topology) that the converse fails to hold. A rather nontrivial theorem is the following:

Theorem

A path-connected Hausdorff space $X$ is arc-connected.

This immediately generalizes to the statement that in a Hausdorff space $X$, any two points that can be connected by a path $\alpha: I \to X$ can be connected by an arc: just apply the theorem to the image $\alpha(I)$.

For a proof of this theorem, see Willard, theorem 31.2. More precisely, that result states that a Peano space, i.e., a compact, connected, locally connected, and metrizable space, is arc-connected if it is path-connected. It then suffices to observe that the continuous image $\alpha(I) \subseteq X$ of a path is in fact a Peano space, so that the path $\alpha: I \to \alpha(I)$ can be replaced by an arc.

Lemma

If $X$ is Hausdorff and there is a continuous surjection $f: I \to X$, then $X$ is a Peano space.

Proof

Obviously $X$ is compact (Hausdorff) and connected. $X$ is a quotient space of $I$, since $f$ is a closed surjection (using compactness of $I$ and Hausdorffness of $X$), and therefore $X$ is locally connected by this lemma. Being compact Hausdorff, $X$ is regular, so to show metrizability it suffices by the Urysohn metrization theorem to show $X$ is second-countable.

Let $\mathcal{B}$ be a countable base for $I$ and let $\mathcal{C}$ be the collection consisting of finite unions of elements of $\mathcal{B}$. We claim $\forall_f(\mathcal{C}) \coloneqq \{\forall_f(C) = \neg f(\neg C): C \in \mathcal{C}\}$ is an (evidently countable) base for $X$. Indeed, suppose $U \subseteq X$ is open and $p \in U$; then $f^{-1}(p)$ is compact, so there exist finitely many $B_1, \ldots, B_n \in \mathcal{B}$ with

$$f^{-1}(p) \subseteq B_1 \cup \ldots \cup B_n \subseteq f^{-1}(U).$$

Put $C = B_1 \cup \ldots \cup B_n$. The first inclusion is equivalent to $p \in \forall_f(C)$ by the adjunction $f^{-1} \dashv \forall_f$. The second inclusion implies $\forall_f(C) \subseteq \forall_f f^{-1}(U) = U$, where the equality $\forall_f f^{-1} = id$, equivalent to $\exists_f f^{-1} = id$, follows from surjectivity of $f$. Thus we have shown $\forall_f(\mathcal{C})$ is a base.

The converse of this lemma is the celebrated Hahn-Mazurkiewicz theorem:

Theorem

Let $X$ be a nonempty Hausdorff space. Then there exists a continuous surjection $\alpha: [0, 1] \to X$ if $X$ is a Peano space. In particular, a nonempty Peano space is path-connected.

(The terminology “Peano space” is given in recognition of Peano’s discovery of space-filling curves, as for example the unit square.)

Path-components functor

As above, let $\pi_0 \colon Top \to Set$ be the functor which assigns to each space $X$ its set of path components $\pi_0(X)$.

Proposition

The functor $\pi_0 \colon Top \to Set$ preserves arbitrary products.

Proof

Let $X_i$ be a family of spaces; we must show that the comparison map

$$\pi_0(\prod_i X_i) \to \prod_i \pi_0(X_i)$$

is invertible. Injectivity: suppose $(x_i), (y_i) \in \prod_i X_i$ are tuples that map to the same tuple of path-components $(c_i)$; we must show that $(x_i)$ and $(y_i)$ belong to the same path component. For each $i$, both $x_i$ and $y_i$ belong to $c_i$, so we may choose a path $\alpha_i: I \to X_i$ connecting $x_i$ to $y_i$. Then $\langle \alpha_i \rangle \colon I \to \prod_i X_i$ connects $(x_i)$ to $(y_i)$. (Note this uses the axiom of choice.) Surjectivity: for any tuple $(c_i) \in \prod_i \pi_0(X_i)$, the component $c_i$ is nonempty for each $i$, so we may choose an element $x_i$ therein. Then $(x_i)$ maps to $(c_i)$. Again this uses the axiom of choice.

An elegant proof of the previous proposition but for preservation of finite products is as follows: both $\hom(I, -)$ and $\hom(1, -)$ preserve products, and a reflexive coequalizer of product-preserving functors $C \to Set$, being a sifted colimit, is also product-preserving.

Proposition

The functor $\pi_0 \colon Top \to Set$ preserves arbitrary coproducts.

Proof

The functor $\hom(I, -) \colon Top \to Set$ preserves coproducts since $I$ is connected, and similarly for $\hom(1, -)$. The coequalizer of a pair of natural transformations between coproduct-preserving functors is also a coproduct-preserving functor.

Pseudo-arcs

Point-set topology is filled with counterexamples. An unusual type of example is that of pseudo-arc:

Definition

A pseudo-arc is a metric continuum with more than one point such that every subcontinuum (a subspace that is a continuum) cannot be expressed as a union of two proper subcontinua.

A pseudo-arc $X$ is necessarily totally path-disconnected: two distinct points $x, y$ of $X$ cannot be connected by a path in $X$. Indeed, the image of such a# path would be a path-connected Hausdorff space, hence arc-connected by Theorem . Letting $\alpha: [0, 1] \to X$ be an arc from $x$ to $y$, we have that the continuum $\alpha([0, 1])$ is a union of proper subcontinua $\alpha([0, 1/2])$ and $\alpha([1/2, 1])$, a contradiction. Thus, a pseudo-arc is an example of a compact connected metrizable space that is totally path-disconnected.

Remarkably, all pseudo-arcs are homeomorphic, and a pseudo-arc is a homogeneous space. Perhaps also remarkable is the fact that the collection of pseudo-arcs in the Hilbert cube $Q$ (or in any Euclidean space) is a dense $G_\delta$ set (see G-delta set) in the Polish space of all nonempty compact subsets of $Q$ under the Hausdorff metric; see Bing2, theorem 2.

A typical way in which pseudo-arcs arise is through inverse limits of dynamical systems. One of the original constructions is due to Henderson:

Theorem

There is a $C^\infty$ function $f: I \to I$ such that the limit of the diagram

$$\ldots \stackrel{f}{\to} I \stackrel{f}{\to} I \stackrel{f}{\to} I$$

is a pseudo-arc.

Roughly speaking, Henderson’s $f$ is a small “notched” perturbation of the squaring function $[0, 1] \to [0, 1]: x \mapsto x^2$, as illustrated on page 38 (of 58) here.

Properties

Corollary

(intermediate value theorem)

Regard the real numbers $\mathbb{R}$ with their Euclidean metric topology, and consider a closed interval $[a,b] \subset \mathbb{R}$ equipped with its subspace topology.

Then a continuous function

$$f \colon [a,b] \longrightarrow \mathbb{R}$$

takes every value in between $f(a)$ and f(b).

Proof

By example the interval $[a,b]$ is connected. By example also its image $f([a,b]) \subset \mathbb{R}$ is connected. By example that image is hence itself an interval. This implies the claim.

Proposition

(topological closure of connected subspace is connected)

Let $(X,\tau)$ be a topological space and let $S \subset X$ be a subset which, as a subspace, is connected. Then also the topological closure $Cl(S) \subset X$ is connected.

Proof

Suppose that $Cl(S) = A \sqcup B$ with $A,B \subset X$ disjoint open subsets. We need to show that one of the two is empty.

But also the intersections $A \cap S\,,B \cap S \subset S$ are disjoint subsets, open as subsets of the subspace $S$ with $S = (A \cap S) \sqcup (B \cap S)$. Hence by the connectedness of $S$, one of $A \cap S$ or $B \cap S$ is empty. Assume $B \cap S$ is empty, otherwise rename. Hence $A \cap S = S$, or equivalently: $S \subset A$. By disjointness of $A$ and $B$ this means that $S \subset Cl(S) \setminus B$. But since $B$ is open, $Cl(S) \setminus B$ is still closed, so that

$$(S \subset Cl(S) \setminus B) \Rightarrow (Cl(S) \subset Cl(S) \setminus B) \,.$$

This means that $B = \emptyset$, and hence that $Cl(S)$ is connected.

Proposition

(connected components are closed)

Let $(X,\tau)$ be a topological space. Then its connected components (def. ) are closed subsets.

Proof

By definition, the connected components are maximal elements in the set of connected subspaces pre-ordered by inclusion. By prop. this means that they must contain their closures, hence they must equal their closures.

Remark

Prop. implies that when a space has a finite set of connected components, then they are not just closed but also open, hence clopen subsets (because then each is the complement of a finite union of closed subspaces).

For a non-finite set of connected components this remains true if the space is locally connected. See this prop.

Related concepts

• locally connected topological space

• connected topos, locally connected topos

• cohesive topos

• contractible topological space

• almost connected topological group

• A space in which the only connected subspaces are the singletons and the empty set is called totally disconnected space.

• The connected subsets of a space form a connectology.

• simply connected topological space

• simple topological space

• nilpotent topological space

• compact connected space

References

Examples of countable connected Hausdorff spaces were give in

• R.H. Bing, A connected countable Hausdorff space, Proc. Amer. Math. Soc. 4 (1953), 474.

• Solomon W. Golomb, A Connected Topology for the Integers, Amer. Math. Monthly, Vol. 66 No. 8 (Oct. 1959), 663-665.

Discussion of connectedness in constructive mathematics is in

• Paul Taylor, Def. 13.2 of: A lambda calculus for real analysis, In: Journal of Logic & Analysis 2 5 (2010) 1–115 $[$L&A:63/25, pdf, webpage$]$

Related discussion in cohesive homotopy type theory:

• Mike Shulman, Thm. 11.1, 11.3 Brouwer’s fixed-point theorem in real-cohesive homotopy type theory, Mathematical Structures in Computer Science 28 6 (2018) 856-941 $[$arXiv:1509.07584, doi:10.1017/S0960129517000147$]$

Material on arc-connected spaces and the Hahn-Mazurkiewicz theorem can be found in Chapter 31 of

• Stephen Willard, General Topology, Addison-Wesley 1970. (online)

Material on pseudo-arcs can be found in

• R.H. Bing, Concerning hereditarily indecomposable continua, Pacific J. Math. Volume 1, Number 1 (1951), 43-51. (Project Euclid)