nLab
connected space

Context

Topology

topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory

Introduction

Basic concepts

Universal constructions

Extra stuff, structure, properties

Examples

Basic statements

Theorems

Analysis Theorems

topological homotopy theory

t

Contents

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 of 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 spaces; see below).

Definitions

Elementary definition

Definition

(connected topological space)

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

  1. For all pairs of topological spaces (X 1,τ 1),(X 2,τ 2)(X_1, \tau_1), (X_2, \tau_2) such that (X,τ)(X, \tau) is homeomorphic to their disjoint union space

    (X,τ)(X 1,τ 1)(X 2,τ 2) (X,\tau) \simeq (X_1,\tau_1) \sqcup (X_2,\tau_2)

    then exactly one of the two spaces is the empty space.

  2. For all pairs of open subsets U 1,U 2XU_1, U_2 \subset X if

    U 1U 2=XAandAU 1U 2= 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 COXCO \subseteq X is clopen (both closed and open), then CO=X CO = X if and only if COCO is inhabited.

Remark

According to def. 1 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 first item of def. 1 the “exactly one” to “at least one” and in the second item “if and only if” to “if”.

Proposition

The conditions in def. 1 are indeed equivalent.

Proof

First consider the equivalence of the first two statements:

Suppose that in every disjoint union decomposition of (X,τ)(X,\tau) then exactly one summand is empty. Now consider two disjoint open subsets U 1,U 2XU_1, U_2 \subset X whose union is XX and whose intersection is empty. We need to show that exactly one of the two subsets is empty.

Write (U 1,τ 1)(U_1, \tau_{1}) and (U 2,τ 2)(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(U 1,τ 1)(U 2,τ 2) X \simeq (U_1, \tau_1) \sqcup (U_2, \tau_2)

because by assumption every open subset UXU \subset X is the disjoint union of open subsets of U 1U_1 and U 2U_2, respectively:

U=UX=U(U 1U 2)=(UU 1)(UU 2), 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 2UU_1, U_2 \subset U with U 1U 2=XU_1 \cup U_2 = X and U 1U 2=U_1 \cap U_2 = \emptyset then exactly one of the two is empty. Now consider a homeomorphism of the form (X,τ)(X 1,τ 1)(X 2,τ 2)(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 2XX_1, X_2 \subset X are disjoint open subsets of XX which cover XX. 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 COXCO \subset X is both closed and open, this means equivalently that it is open and that its complement XCOX \setminus CO is also open, hence equivalently that there are two open subsets CO,X\COXCO, X \backslash CO \subset X whose union is XX and whose intersection is empty. This way the third condition is equivalent to the second.

Category-theoretic definition

In the language of category theory def. 1 may be rephrased as follows:

Write Top for the category of all topological space.

Then a topological space XX is connected precisely if the representable functor

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

preserves coproducts.

It is equivalent to just require that it preserves binary coproducts (a detailed proof 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)hom(X,Y+Z), hom(X,Y) + hom(X,Z) \to hom(X,Y + Z) ,

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

The variant of the definition according to remark 1, which regards the empty space means in terms of category theoretic language that one requires only that the maps

hom(X,Y)+hom(X,Z)hom(X,Y+Z) 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 (much as one disqualifies 11 when one defines the notion of prime number). See also the discussion at empty space and too simple to be simple.

Connected components

Every topological space XX admits an equivalence relation \sim where xyx \sim y means that xx and yy belong to some subspace which is connected. The equivalence class Conn(x)Conn(x) of an element xx is thus the union of all connected subspaces containing xx; it follows readily from Result 4 that Conn(x)Conn(x) is itself connected. It is called the connected component of xx. It is closed, by Result 5. 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,τ)(X,\tau) a topological space, then its connected components are the equivalence classes under the equivalence relation on XX which regards two points as equivalent if they both sit in some open subset which, as a topological subspace (example \ref{SubspaceTopology}), is connected (def. \ref{ConnectedTopologicalSpace}):

(xy)(UX\open((x,yU)AandA(Uis connected))). (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) \,.

There is another equivalence relation q\sim_q where x qyx \sim_q y if f(x)=f(y)f(x) = f(y) for every continuous f:XDf: X \to D mapping to a discrete space DD. The equivalence class of xx may be alternatively described as the intersection of all clopens that contain xx. This is called the quasi-component of xx, denoted here as QConn(x)QConn(x). It is easy to prove that

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

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

Example

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

X={(0,0),(0,1)} n1{1/n}×[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)}Conn((0, 1)) = \{(0, 1)\}, but QConn((0,1))={(0,0),(0,1)}QConn((0, 1)) = \{(0, 0), (0, 1)\}.

Examples

Basic examples

Example

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

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

Proof

By definition of locally constant functions, every point xXx \in X has an open neighbourhood U xU_x such that the restriction f| U xf\vert_{U_x} is a constant function. The unions of these neighbourhoods for a fixed constant value hence are disjoint open subsets that constitute a cover of XX. By connectedness this cover must consist of a single non-empty element. But by construction this mans that ff is constant.

Example

(continuous images of connected spaces are connected)

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

Let XX be a connected topological space, let YY be any topological space, and let

f:XY f \;\colon\; X \longrightarrow Y

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

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

for f(X)f(X) equipped either with he subspace topology relative to YY or the quotient topology relative to XX. In either case:

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

Proof

Let U 1,U 2f(X)U_1,U_2 \subset f(X) be two open subsets such that U 1U 2=f(X)U_1 \cup U_2 = f(X) and U 1U 2=U_1 \cap U_2 = \emptyset. We need to show that precisely one of them is the empty set.

Since pp is a continuous function, also the pre-images p 1(U 1),p 1(U 2)Xp^{-1}(U_1), p^{-1}(U_2) \subset X are open subsets and are still disjoint. Since pp is surjective it also follows that p 1(U 1)p 1(U 2)=Xp^{-1}(U_1) \cup p^{-1}(U_2) = X. Since XX is connected, it follows that one of these two pre-images is the empty set. But again sicne pp is surjective, this implies that precisely one of U 1,U 2U_1, U_2 is empty, which means that f(X)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 SetSet commute with wide pullbacks. More memorably: connected colimits of connected spaces are connected.

Example

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

(see also prop. 3 below)

Example

(product space of connected spaces is connected)

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

Proof

This relies on some special features of Top An 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 2iIX iU_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 iX_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 1U_1 and U 2U_2 are non-empty.

Observe first that if so, then we could find x 1U 1x_1 \in U_1 and x 2U 2x_2 \in U_2 whose coordinates differed only a a finite subset of II. This is since by the nature of the Tychonoff topology π i(U 1)=X i\pi_i(U_1) = X_i and π i(U 2)=X i\pi_i(U_2) = X_i for all but a finite number of iiIi \in iI.

Next observe that we then could even find x 1U 1x'_1 \in U_1 that differed only in a single coordinate from x 2x_2: Because pick one coordinate in which x 1x_1 differs from x 2x_2 and change it to the corresponding coordinate of x 2x_2. Since U 1U_1 and U 2U_2 are a cover, the resulting point is either in U 1U_1 or in U 2U_2. If it is in U 2U_2, then x 1x_1 already differed in only one coordinate from x 2x_2 and we may take x 1x 1x'_1 \coloneqq x_1. If instead the new point is in U 1U_1, then rename it to x 1x_1 and repeat the argument. By induction this finally yields an x 1x'_1 as claimed.

Therefore it is now sufficient to see that it leads to a contradiction to assume that there are points x 1U 1x_1 \in U_1 and x 2U 2x_2 \in U_2 that differ in only the i 0i_0th coordinate, for some i 0Ii_0 \in I then x 1=x 2x_1 = x_2.

Observe that the inclusion

ι:X i 0iIX i \iota \colon X_{i_0} \longrightarrow \underset{i \in I}{\prod} X_i

which is the identity on the i 0i_0th component and is otherwise constant on the iith component of x 1x_1 or equivalently of x 2x_2 is a continuous function, by the nature of the Tychonoff topology.

Therefore also the restrictions ι 1(U 1)\iota^{-1}(U_1) and ι 1(U 2)\iota^{-1}(U_2) are open subsets. Moreover they are still disjoint and cover X iX_i. Hence by the connectedness of X iX_i, precisely one of them is empty. This means that the i 0i_0-component of both x 1x_1 and x 2x_2 must be in the other subset of X iX_i, and hence that x 1x_1 and x 2x_2 must both be in U 1U_1 or both in U 2U_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 SS \subset \mathbb{R} is connected (def. 1) precisely if it is an interval, hence precisely if

x,ySr((x<r<y)(rS)). \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} iI\{ I_i \subset \mathbb{R} \}_{i \in I} a set of disjoint intervals, then II is the set of connected components of the union iII i\underset{i \in I}{\cup} I_i.

Proof

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

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

But since x<rx \lt r and r<yr \lt y both these open subsets were non-empty, thus contradicting the assumption that SS 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)×:y0}Q = \{(x, y) \in \mathbb{Q} \times \mathbb{Q}: y \geq 0\}, topologized by defining a basis of neighborhoods N ϵ,a,bN_{\epsilon, a, b} for each point (a,b)Q(a, b) \in Q and ϵ>0\epsilon \gt 0:

N a,b{(a,b)}{(s,0)Q:|a+b/θs|<ϵ}{(s,0)Q:|ab/θs|<ϵ}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 θ<0\theta \lt 0 is some chosen fixed irrational number. It is easy to see this space is Hausdorff (using the fact that θ\theta is irrational). However, the closure of N ϵ,a,bN_{\epsilon, a, b} consists of points (x,y)(x, y) of Q×QQ \times Q with either (xa)ϵ(yb)/θ(xa)+ϵ(x-a) - \epsilon \leq (y-b)/\theta \leq (x-a) + \epsilon or (xa)ϵ(yb)/θ(xa)+ϵ(x-a) - \epsilon \leq -(y-b)/\theta \leq (x-a) + \epsilon, in other words, the union of two infinitely long strips of width 2ϵ2\epsilon and slopes θ\theta, θ-\theta. Clearly any two such closures intersect, and therefore the space is connected.

Example t

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{ak+b|k=1,2,}A_{a,b} \coloneqq \{a k + b | k = 1,2, \ldots\}, where a,ba, 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.

Connected components

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 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).

Path-connectedness

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

Definition

(continuous path in a topological space)

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

γ:[0,1](X,τ) \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 γ(0)X\gamma(0) \in X with the point γ(1)X\gamma(1) \in X.

For xXx \in X a fixed point, then the subset

PConn x(X){yX|pathγ((γ(0)=x)andA(γ(1)=y))} 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 xx.

The set of path connected components of XX is denoted

π 0(X). \pi_0(X) \,.

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

hom([0,1],X)ev 1ev 0hom(1,X)π 0(X). \hom([0, 1], X) \stackrel{\overset{ev_0}{\to}}{\underset{ev_1}{\to}} \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)hom([0,1],X)\hom(!, X): \hom(1, X) \to \hom([0, 1], X).

(We can even topologize π 0(X)\pi_0(X) by taking the coequalizer in TopTop of

X [0,1]ev 1ev 0X,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][0, 1] is exponentiable. The resulting quotient space will be discrete if XX is locally path-connected.)

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

It follows easily from the basic results above that:

Lemma

A path connected space XX is connected.

Proof

Assume it were not, then it would be covered by two disjoint inhabited open subsets U 1,U 2XU_1, U_2 \subset X. But by path connectedness there were a continuous path γ:[0,1]X\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 γ 1(U 1),γ 1(U 2)[0,1]\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 xx); see the following example. The path components and connected components do coincide if XX is locally path-connected.

Example

The topologist’s sine curve

{(x,y) 2:(0<x1y=sin(1/x))(0=x1y1)} \{ (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)y = \sin(1/x).)

The basic categorical Results 3, 4, and 6 above carry over upon replacing “connected” by “path-connected”. (As of course does example 7, 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 XX is arc-connected if for any two distinct x,yXx, y \in X there exists an injective continuous map α:IX\alpha: I \to X such that α(0)=x\alpha(0) = x and α(1)=y\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 XX is arc-connected.

This immediately generalizes to the statement that in a Hausdorff space XX, any two points that can be connected by a path α:IX\alpha: I \to X can be connected by an arc: just apply the theorem to the image α(I)\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 α(I)X\alpha(I) \subseteq X of a path is in fact a Peano space, so that the path α:Iα(I)\alpha: I \to \alpha(I) can be replaced by an arc.

Lemma

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

Proof

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

Let \mathcal{B} be a countable base for II and let 𝒞\mathcal{C} be the collection consisting of finite unions of elements of \mathcal{B}. We claim f(𝒞){ f(C)=¬f(¬C):C𝒞}\forall_f(\mathcal{C}) \coloneqq \{\forall_f(C) = \neg f(\neg C): C \in \mathcal{C}\} is an (evidently countable) base for XX. Indeed, suppose UXU \subseteq X is open and pUp \in U; then f 1(p)f^{-1}(p) is compact, so there exist finitely many B 1,,B nB_1, \ldots, B_n \in \mathcal{B} with

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

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

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

Theorem

Let XX be a nonempty Hausdorff space. Then there exists a continuous surjection α:[0,1]X\alpha: [0, 1] \to X if XX 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 π 0:TopSet\pi_0 \colon Top \to Set be the functor which assigns to each space XX its set of path components π 0(X)\pi_0(X).

Proposition

The functor π 0:TopSet\pi_0 \colon Top \to Set preserves arbitrary products.

Proof

Let X iX_i be a family of spaces; we must show that the comparison map

π 0( iX i) iπ 0(X i)\pi_0(\prod_i X_i) \to \prod_i \pi_0(X_i)

is invertible. Injectivity: suppose (x i),(y i) iX i(x_i), (y_i) \in \prod_i X_i are tuples that map to the same tuple of path-components (c i)(c_i); we must show that (x i)(x_i) and (y i)(y_i) belong to the same path component. For each ii, both x ix_i and y iy_i belong to c ic_i, so we may choose a path α i:IX i\alpha_i: I \to X_i connecting x ix_i to y iy_i. Then α i:I iX i\langle \alpha_i \rangle \colon I \to \prod_i X_i connects (x i)(x_i) to (y i)(y_i). (Note this uses the axiom of choice.) Surjectivity: for any tuple (c i) iπ 0(X i)(c_i) \in \prod_i \pi_0(X_i), the component c ic_i is nonempty for each ii, so we may choose an element x ix_i therein. Then (x i)(x_i) maps to (c i)(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,)\hom(I, -) and hom(1,)\hom(1, -) preserve products, and a reflexive coequalizer of product-preserving functors CSetC \to Set, being a sifted colimit, is also product-preserving.

Proposition

The functor π 0:TopSet\pi_0 \colon Top \to Set preserves arbitrary coproducts.

Proof

The functor hom(I,):TopSet\hom(I, -) \colon Top \to Set preserves coproducts since II is connected, and similarly for hom(1,)\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 XX is necessarily totally path-disconnected: two distinct points x,yx, y of XX cannot be connected by a path in XX. Indeed, the image of such a# path would be a path-connected Hausdorff space, hence arc-connected by Theorem 1. Letting α:[0,1]X\alpha: [0, 1] \to X be an arc from xx to yy, we have that the continuum α([0,1])\alpha([0, 1]) is a union of proper subcontinua α([0,1/2])\alpha([0, 1/2]) and α([1/2,1])\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 QQ (or in any Euclidean space) is a dense G δG_\delta set (see G-delta set) in the Polish space of all nonempty compact subsets of QQ 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 C^\infty function f:IIf: I \to I such that the limit of the diagram

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

is a pseudo-arc.

Roughly speaking, Henderson’s ff is a small “notched” perturbation of the squaring function [0,1][0,1]:xx 2[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][a,b] \subset \mathbb{R} equipped with its subspace topology.

Then a continuous function

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

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

Proof

By example 7 the interval [a,b][a,b] is connected. By example 3 also its image f([a,b])f([a,b]) \subset \mathbb{R} is connected. By example 7 that image is hence itself an interval. This implies the claim.

Proposition

(topological closure of connected subspace is connected)

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

Proof

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

But also the intersections AS,BSSA \cap S\,,B \cap S \subset S are disjoint subsets, open as subsets of the subspace SS with S=(AS)(BS)S = (A \cap S) \sqcup (B \cap S). Hence by the connectedness of SS, one of ASA \cap S or BSB \cap S is empty. Assume BSB \cap S is empty, otherwise rename. Hence AS=SA \cap S = S, or equivalently: SAS \subset A. By disjointness of AA and BB this means that SCl(S)BS \subset Cl(S) \setminus B. But since BB is open, Cl(S)BCl(S) \setminus B is still closed, so that

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

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

Proposition

(connected components are closed)

Let (X,τ)(X,\tau) be a topological space. Then its connected components (def. 2) are closed subsets.

Proof

By definition, the connected components are maximal elements in the set of connected subspaces pre-ordered by inclusion. By prop. 3 this means that they must contain their closures, hence they must equal their closures.

Remark

Prop. 4 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.

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.

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)

Revised on September 21, 2017 09:02:16 by Todd Trimble (47.18.191.229)