Connected Space
A connected topological space is a space that cannot be expressed as a union of two disjoint open subsets. Connectedness is a property that helps to classify and describe topological spaces; it is also an important assumption in many important applications, including the intermediate value theorem.
As with compactness, the formal definition of connectedness is not exactly the most intuitive. The more natural definition is that a space is connected if a curve can be drawn between any two points in the space ("connecting" the points). With the appropriate assumptions and definition, this down-to-earth definition becomes an important alternate notion of connectedness known as path-connectedness. Connected spaces are not always path-connected (although the converse is true).
Contents
Definitions
Let \(X\) be a topological space. Then \(X\) is said to be disconnected if \(X = U_1 \cup U_2,\) with \(U_1,U_2\) open subsets of \(X\) such that \( U_1 \cap U_2 = \varnothing.\) Otherwise, \(X\) is connected.
An equivalent formulation of this definition involves clopen sets, defined as follows:
A clopen set is a set that is both closed and open.
Clearly, \(X\) and \(\varnothing\) are clopen, and the definition of connectedness is related to other clopen sets:
\(X\) is connected if and only if the only clopen subsets of \(X\) are \(X\) and \(\varnothing.\)
This is clear, since if \(U_1,U_2\) are open subsets with \( U_1 \cup U_2 = X\) and \( U_1 \cap U_2 = \emptyset,\) then \(U_1\) and \(U_2\) are clopen. (This reasoning works in reverse as well.)
Subspaces
Let \(Y\) be a subset of a topological space \(X.\) Recall that \(Y\) inherits a topology from \(X\) called the subspace topology, where the open sets of \(Y\) are the intersections of open sets of \(X\) with \(Y.\) Then a subset \(Y\) of \(X\) is connected if and only if it is connected as a topological space, with the subspace topology.
Consider the subset \( Y = [0,1) \cup (1,2] \subseteq \mathbb R.\) Then \(Y\) is disconnected, because it is the union of \( [0,1) \) and \( (1,2].\) Neither of these are open subsets of \( \mathbb R,\) but they are open subsets of \(Y\) with the subspace topology \(\big(\)since they are the intersection of e.g. \( (-1,1)\) and \((1,3)\) with \(Y\big).\)
Any interval in \( \mathbb R\) is connected.
All proofs of this result use some form of the completeness property of \(\mathbb R.\) Here is one such proof. For simplicity's sake, we will only give the proof when the interval is closed, and without loss of generality the interval can be assumed to be \([0,1]\) in this case. The result for general intervals follows quickly from the result for closed intervals, by embedding a non-closed interval inside a closed interval with the same endpoints (left as an exercise for the reader).So suppose \( [0,1] = U \cup V\) with \(U,V\) open, disjoint, nonempty. Without loss of generality, pick \( a\in U, b \in V\) with \( 0 < a < b < 1.\) Let \[ c = \sup \big( \{ x \in U, x < b\} \big). \] (This is where completeness is used, to know that this supremum exists.)
If \(c \in U,\) then find an interval \( (c-\epsilon,c+\epsilon)\) contained entirely in \(U\) \((\)this is possible since \(U\) is open\().\) Note that \(b\) is not in this interval because \(b \notin U.\) Since \(c\le b,\) it follows that in this case \(c+\epsilon \le b.\) This violates the description of \(c\) as being larger than all elements of \(U\) which are \( \le b,\) since it is not larger than e.g. \(c+\frac{\epsilon}{2}.\)
If \(c \in V,\) then a similar argument shows that \(c\) is not the least upper bound for the set \( \{x \in U, x < b\}.\) So in both cases there is a contradiction. \(_\square\)
Connected Components
Every topological space \(X\) can be partitioned into connected components; these are simply the maximal connected subsets of \(X,\) ordered by inclusion. The connected components of \(X\) are non-empty, disjoint, and their union is \(X.\) One way to see this is by considering the equivalence relation on points of \(X\) defined by \(x \sim y \) if and only if \(x\) and \(y\) are both contained in a connected set; then the connected components are the equivalence classes under this relation.
The fact that this relation is transitive requires the following important result:
Let \(X\) be a topological space and \(Y, Z\) two connected subspaces. Then if \(Y \cap Z\) is non-empty, \(Y \cup Z\) is connected.
Suppose \(Y \cup Z\) is disconnected, e.g. \( Y\cup Z = U \cup V, \) where \(U,V\) are disjoint, nonempty, and open in \( Y\cup Z.\) Then pick a point \(x \in Y \cap Z.\) Suppose without loss of generality that \(x \in U.\) So \(U \cap Y\) is nonempty, and is open in \(Y.\) Now \(V \cap Y\) is open in \(Y,\) and disjoint from \(U \cap Y,\) and \(U \cap Y\) and \(V \cap Y\) cover \(Y.\) Since \(Y\) is connected, the only possibility is that \(V \cap Y = \varnothing.\) So \(Y \subseteq U\) and \(V \subseteq Z.\) But now we can use that \(Z\) is connected: \(U \cap Z\) and \(V\) are disjoint and nonempty and cover \(Z,\) which is impossible. This contradiction shows that \(Y \cup Z\) must be connected. \(_\square\)
Path-connectedness
This section describes a more straightforward definition of a connectedness property, involving joining points by paths inside the topological space. First, it is necessary to define a path.
Let \(X\) be a topological space. A path between two points \(x,y\in X\) is a continuous function \( f \colon [0,1] \to X\) such that \(f(0) = x\) and \(f(1) = y.\)
A topological space \(X\) is path-connected if there is a path between any two points in \(X.\)
A path-connected topological space is connected.
Suppose \(X\) is path-connected but not connected. So \(X = U \cup V \) with \(U,V\) nonempty disjoint open sets. Take \(x \in U,\) \(y \in V,\) and let \(f\) be the path between them. Then \( f^{-1}(U)\) and \( f^{-1}(V)\) are disjoint, nonempty, open subsets of \([0,1]\) which cover \( [0,1].\) But this is impossible, since a closed interval in \(\mathbb R\) is connected. \(_\square\)
The converse of this theorem is false. Here is one well-known counterexample in \({\mathbb R}^2,\) called the "topologist's sine curve."
Let \(X\) be the subset of \({\mathbb R}^2\) consisting of the points \(\big(x,\sin \frac1x \big)\) for all \(x > 0,\) plus the origin \((0,0).\) Show that \(X\) is connected but not path-connected.
Note that \(X \setminus \{(0,0)\}\) is path-connected, since it is the graph of a continuous function. Suppose \(X\) is not connected; then an expression of \(X\) as a disjoint union of \(U\) and \(V,\) with \((0,0)\in U,\) leads to an expression of \(X\setminus \{(0,0)\}\) as a disjoint union of \(U\setminus \{(0,0)\}\) and \(V.\) The key point is that \((0,0)\) is a limit point of \(X,\) so \(U\setminus (0,0)\) is nonempty. This is a contradiction, since \(X \setminus \{(0,0)\}\) is path-connected and hence connected.
To see that \(X\) is not path-connected, the idea is to show that any path starting at \((0,0)\) and staying inside \(X\) is the trivial path \(\big(\)which stays on \((0,0)\big).\) The intuition is, roughly, that a path has no way to "jump" from the origin onto the rest of \(X,\) because \(X\) is "infinitely wavy" near \(x=0.\) To make this formal, suppose \(f \colon [0,1] \to X\) is a path with \(f(0) = (0,0).\) Suppose it is not the trivial path. Let \[ t_0 = \inf\big(\{t \in [0,1] : f(t) \ne (0,0)\}\big). \] By continuity of \(f,\) \(f(t_0) = \lim\limits_{t \to t_0^-} f(t) = (0,0).\) Also by continuity, there is a \(\delta > 0\) such that the \(y\)-coordinates of \(f(t)\) for \(t \in (t_0,t_0+\delta)\) are less than, say, \(\frac12.\) Because \(t_0\) is the infimum, there is some point \(t_1\) in that interval for which \(f(t_1) \ne (0,0).\) But the interval \((t_0,t_1)\) must contain infinitely many points whose \(y\)-coordinate is \(1\) (the "infinitely wavy" part of the argument), which is a contradiction. \(_\square\)
Properties of Connected Sets
Continuous images: If \(f \colon X \to Y\) is a continuous function, and \(X\) is connected, then \(f(X)\) is connected.
This is a generalization of the argument that path-connected implies connected: if \(f(X) = U \cup V\) with \(U,V\) nonempty, open, disjoint, then \(f^{-1}(U)\) and \(f^{-1}(V)\) are nonempty, open, disjoint sets whose union is \(X,\) which is impossible since \(X\) is connected.
The same result holds for path-connected sets: the continuous image of a path-connected set is path-connected.
The proof is immediate: if \(p \colon [0,1] \to X\) is a path connecting \(x\) to \(y,\) then \(f \circ p \colon [0,1] \to Y\) is a path connecting \(f(x)\) to \(f(y).\)
A special case of this result is the intermediate value theorem: if \(f\colon [a,b] \to \mathbb R \) is continuous and \( y\) is any real number between \(f(a)\) and \(f(b),\) then there is a \(c \in [a,b] \) such that \(f(c) = y.\) This is because the image of \(f\) is path-connected, and hence connected. If \( y \notin f\big([a,b]\big),\) then the sets \( \big\{ (-\infty,y) \cap f([a,b]) \big\} \) and \( (y,\infty) \cap f\big([a,b]\big)\) are open, disjoint, nonempty sets whose union is \( f\big([a,b]\big),\) which is a contradiction.
Closure: The closure of a connected set is connected \((\)in fact, any set \(Y\) such that \(X \subseteq Y \subseteq {\overline X}\) is connected\().\) This is a generalization of the topologist's sine curve argument: adding limit points does not affect connectedness.
Unions and intersections: The union of two connected sets is connected if their intersection is nonempty, as proved above. But if their intersection is empty, the union may not be connected \((\)e.g. two disjoint open intervals in \(\mathbb R).\)
The intersection of two connected sets is not always connected.
Show that no two of the sets \( (0,1), [0,1),\) and \([0,1]\) are homeomorphic.
Deleting any point from \( (0,1) \) gives a disconnected space. There is exactly one point in \( [0,1) \) which can be deleted to give a connected space. There are exactly two points in \( [0,1]\) which can be deleted to give a connected space. If there were a homeomorphism between \( [0,1)\) \(\big(\)or \( [0,1]\big)\) and \( (0,1),\) then the image of \( (0,1) \) would be \( (0,1) \) with one (or two) points deleted, which contradicts the fact that continuous images of connected sets are connected. Similarly, if there were a homeomorphism \( [0,1] \to [0,1),\) the image of \( (0,1) \) would be \( [0,1)\) with two points deleted, which cannot be connected. \(_\square\)
References
- Stella, B. Union et intersection d'ensembles. Retrieved October 30, 2006, from https://en.wikipedia.org/wiki/Connected_space#/media/File:Union_et_intersection_d%27ensembles.svg