Connectivity#
This section introduces connectivity (or connectedness). Connectivity is a topological property that describes whether a space, or subspace, consists of “one piece.” There are different variants of connectivity. We introduce these variants here and discuss some of their properties.
Definition of connectivity#
The first definition of connectivity amounts to whether a space can be split into a disjoint union of two (non-empty) open sets.
Definition 101 (Connected space)
A topological space \(X\) is disconnected if \(X\) can be written as \(A \cup B,\) where \(A\) and \(B\) are disjoint, non-empty open subsets of \(X.\) We say \(A\) and \(B\) disconnect \(X.\) A space is connected if it is not disconnected.
Some additional intuition on connected spaces
We give some additional intuition on the conditions for connectivity.
Disjoint sets: We require that the sets are disjoint because if not, then our notion of connectivity would not make sense: if we did not require the sets are disjoint, then pairs of sets like \((-\infty, 1)\) and \((-1, \infty)\) would disconnect \(\mathbb{R},\) which would not align with our expectation of what disconnected means.
Non-empty sets: We also require that both sets are non-empty because we can always express a space as the disjoint union of itself and the empty set, both of which are empty. Allowing this would trivialise the definition.
Open sets: Finally, we require that the sets are both open. If we did not require this, then \((-\infty, 0)\) and \([0, \infty)\) would disconnect \(\mathbb{R}.\) Intuitively, this final requirement ensures that neither of the two sets contains a shared boundary: if one set contained elements that are part of the boundary of the other, then this would not align with our expectation of what disconnected means.
An additional observation: We also note the following point about disconnected sets: if \(A, B \subseteq X\) disconnect \(X,\) then both \(A, B\) are open and closed in \(X.\) To see this, note that \(A \cap B = \emptyset\) and \(A \cup B = X.\) Therefore, \(X \setminus A = B\) so the complement of \(A\) in \(X\) is open, which means \(A\) is closed in \(X.\) A similar argument shows \(B\) is closed in \(X\) as well. In this sense, we can consider \(A\) and \(B\) as disconnected or separate. for example, any sequence that converges in \(A\) (or \(B\)) does not have its limit in \(B\) (or \(A\)). We can also go the other way and observe that for any connected space \(X,\) the only subspace of it that is both open and closed is the space \(X\) itself: if \(X\) contains a non-empty subset \(A\) that is both open and closed, then \(B = X \subset A\) is an open subset of \(X\) that is disjoint from \(A,\) and if \(B\) were non-empty then \(A\) and \(B\) would disconnect \(X,\) so \(B = \emptyset\) and \(A = X.\) With this final bit of intuition, whenever a space can be disconnected into \(A\) and \(B,\) we can regard \(A\) and \(B\) as separate sets in \(X,\) whose union does not alter topological properties such as convergence of sequences when we consider the two sets together. See also exercise 2.6.
We can also consider whether a subspace of a topological space is connected, according to the following definition.
Definition 102 (Connected subspace)
Given a subset \(A\) of a topological space \(X,\) we say that \(A\) is connected if \(A\) is connected as a subspace of \(X.\)
The following is a useful equivalent condition for connectedness. For example, in some cases it may be easy to construct a function that helps us show a certain space is disconnected.
Lemma 26 (Equivalent condition for connectedness)
A space \(X\) is disconnected if and only if there exists a contunuous surjective \(f: X \to \{0, 1\}\) with the discrete topology.
Proof: Equivalent condition for connectedness
Suppose \(X\) is disconnected. Then there exist disjoint, non-empty open sets \(A\) and \(B\) such that \(X = A \cup B.\) Define \(f: X \to \{0, 1\}\) by
Now, we note that \(f\) is surjective because \(A\) and \(B\) are non-empty. In addition, \(f\) is continuous because \(f^{-1}(\{0\}) = A\) and \(f^{-1}(\{1\}) = B\) are open. Thus, \(f\) is continuous and surjective.
Conversely, suppose there exists a continuous surjective \(f: X \to \{0, 1\}\) with the discrete topology. Then the disjoint sets \(f^{-1}(\{0\})\) and \(f^{-1}(\{1\})\) are non-empty (because \(f\) is surjective) and open (because \(f\) is continuous). Thus, \(X\) is disconnected.
We now show one of the main results of the section, namely that \([0, 1]\) is connected in \(\mathbb{R}\) with the regular topology.
Theorem 96 (Closed inverval is connected)
The closed interval \([0, 1]\) with the standard topology is connected.
Proof: Closed interval is connected
Suppose \([0, 1]\) is disconnected. Then there exist disjoint, non-empty open sets \(A\) and \(B\) such that \([0, 1] = A \cup B.\) Without loss of generality, we may assume that \(1 \in B.\) Since \(A\) is non-empty, \(\alpha = \sup A\) exists and \(\alpha \in [0, 1].\) There are two possibilities, either \(\alpha \in T\) or \(\alpha \in B.\)
If \(\alpha \in T,\) then \(\alpha < 1\) and we immediately have a contradiction because, since \(A\) is open, there exists an open interval \((\alpha - \varepsilon, \alpha + \varepsilon) \subseteq [0, 1]\) contained in \(A\) for some \(\varepsilon > 0,\) which means that \(\alpha\) cannot be the supremum of \(A.\) Therefore, \(\alpha \in B.\) However, again, this raises a contradiction because, since \(B\) is open, there exists an open interval \((\alpha - \varepsilon, \alpha + \varepsilon) \subseteq [0, 1]\) contained in \(B\) for some \(\varepsilon > 0,\) which means that \(\alpha\) cannot be the supremum of \(A.\) Thus, \([0, 1]\) is connected.
Together with the fact that products of connected spaces are connected, we see that \([0, 1]^n\) is connected, and so the interior of the \(n\)-dimensional disc \(\text{Int}(D^n)\) is also connected.
Lemma 27 (Image of connected space under continuous map is connected)
Let \(f: X \to Y\) be a continuous map. If \(X\) is connected, then \(f(X)\) is too.
Proof: Image of connected space under continuous map is connected
Suppose \(f(X)\) is disconnected. Then there exist disjoint, non-empty open sets \(A\) and \(B\) such that \(f(X) = A \cup B.\) Define \(U = f^{-1}(A)\) and \(V = f^{-1}(B).\) Then \(U\) and \(V\) are disjoint, non-empty open sets in \(X,\) and \(X = U \cup V.\) Therefore \(X\) is disconnected, which is a contradiction. Thus, \(f(X)\) is connected.
We now prove a more general version of a standard theorem in analysis: the intermediate value theorem. Note that our notion of connectivity makes the proof of the intermediate value theorem very short and elegant.
Theorem 97 (Intermediate value theorem)
Suppose \(f: X \to \mathbb{R}\) is continuous and \(X\) is connected. If \(x_0, x_1 \in X\) such that \(f(x_0) < 0 < f(x_1),\) then there exists \(x \in X\) such that \(f(x) = 0.\)
Proof: Intermediate value theorem
Suppose there does not exist \(x \in X\) such that \(f(x) = 0.\) Then \(f(X) \cap (-\infty, 0)\) and \(f(X) \cap (0, \infty)\) are open, non-empty and disjoint sets whose union is \(f(X).\) Therefore, \(f(X)\) is disconnected, which is a contradiction. Thus, there exists \(x \in X\) such that \(f(x) = 0.\)
Path connectivity#
Now we introduce the second notion of connectivity that we will consider, path connectivity. This is a stronger requirement than regular connectivity. It is also perhaps a little more intuitive than regular connectivity: a space is path connected if any two points can be joined by a continuous path, i.e. curve, with the two points as end points. We first define paths.
With paths in place, we are ready to define path connectivity.
Definition 103 (Path connectivity)
A topological space \(X\) is path connected if for all points \(x_0, x_1 \in X,\) there exists a path from \(x_0\) to \(x_1.\)
As mentioned, path connectivity is a stronger notion, i.e. implies, regular connectivity.
Lemma 28 (Path connected implies connected)
If \(X\) is path connected, then \(X\) is connected.
Proof: Path connected implies connected
Let \(X\) be path connected, and let \(f: X \to \{0, 1\}\) be continuous. We want to show that \(f\) is constant.
Let \(x_0, x_1 \in X.\) By path connectedness, there is a map \(\gamma: [0, 1] \to X\) such that \(\gamma(0) = x_0\) and \(\gamma(1) = x_1.\) Then \(f \circ \gamma: [0, 1] \to \{0, 1\}\) is continuous. Since \([0, 1]\) is connected, \(f \circ \gamma\) is constant and \(f(\gamma(0)) = f(\gamma(1)),\) which means that \(f(x_0) = f(x_1).\) Since \(x_0\) and \(x_1\) were arbitrary, \(f\) is constant.
Lemma 29 (Restrictions of homeomorphisms are homeomorphisms)
Let \(f: X \to Y\) be a homeomorphism, and let \(A \subseteq X.\) Then \(f|_A: A \to f(A)\) is a homeomorphism.
Proof: Restrictions of homeomorphisms are homeomorphisms
Since \(f\) is a homeomorphism, it is a bijection, so \(f|_A\) is a bijection. If \(U \subseteq f(A)\) is open, then \(U = f(A) \cap U'\) for some open \(U' \subseteq Y.\) Then \(f^{-1}|_A(U) = f^{-1}(U) \cap A\) is open in \(A\) because \(f^{-1}\) is continuous. Therefore \(f|_A\) is continuous.
Similarly, we can show that \(f|_A^{-1}\) is continuous. If \(V \subseteq A\) is open, then \(V = A \cap V'\) for some open \(V' \subseteq X.\) Then \((f|_A^{-1})^{-1}(V) = (f^{-1})^{-1}(V) = f(V)\) is open in \(f(A)\) because \(f\) is continuous. Therefore \(f|_A^{-1}\) is continuous.
Thus, \(f|_A\) is a homeomorphism.
Components#
We now introduce path components and connected components. These are equivalence classes of points which are path connected or connected, repspectively.
Path components#
Lemma 30 (Path connected equivalence relation)
Let \(X\) be a topological space. For \(x, x' \in X,\) define \(x \sim x'\) if there exists a path from \(x\) to \(x'.\) Then \(\sim\) is an equivalence relation.
Proof: Path connected equivalence relation
Let \(x, x', x'' \in X.\) We show that \(\sim\) satisfies the properties of an equivalence relation.
Reflexivity: Then the constant map \(\gamma: [0, 1] \to X\) defined by \(\gamma(t) = x\) is a path from \(x\) to \(x,\) so \(x \sim x.\)
Symmetry: If \(\gamma: [0, 1] \to X\) is a path from \(x\) to \(x',\) then the map \(\gamma': [0, 1] \to X\) defined by \(\gamma'(t) = \gamma(1 - t)\) is a path from \(x'\) to \(x,\) so \(x \sim x'.\)
Transitivity: If \(\gamma: [0, 1] \to X\) is a path from \(x\) to \(x'\) and \(\gamma': [0, 1] \to X\) is a path from \(x'\) to \(x'',\) then the map \(\gamma'': [0, 1] \to X\) defined by
is a path from \(x\) to \(x'',\) so \(x \sim x''.\)
Definition 104 (Path component)
Equivalence classes of the path connectivity equivalence relation are called path components.
Connected components#
Now we turn our attention to connected components, which are analogous to path components, but for regular connectivity.
Lemma 31 (Union of intersecting connected sets is connected)
Let \(X\) be a topological space. Suppose \(Y_\alpha \subseteq X\) is connected for all \(\alpha \in T,\) and \(\cap_{\alpha \in T} Y_\alpha \neq \emptyset.\) Then \(\cup_{\alpha \in T} Y_\alpha\) is connected.
Proof: Union of intersecting connected sets is connected
Suppose \(Y = \cup{\alpha \in T} Y_\alpha\) is disconnected. Then there exist disjoint, non-empty open sets \(A\) and \(B\) that are subsets of \(Y\) such that \(A \cup B = Y.\) Since \(A\) and \(B\) are open in \(Y,\) there exist open \(A', B' \subseteq X\) such that \(A = A' \cap Y\) and \(B = B' \cap Y.\) Define
Note that \(A_\alpha\) and \(B_\alpha\) are open in \(Y_\alpha.\) Note also that \(A_\alpha \cup B_\alpha = Y_\alpha\) and \(A_\alpha \cap B_\alpha = \emptyset.\) Since \(Y_\alpha\) is connected, we must have either \(A_\alpha = \emptyset\) or \(B_\alpha = \emptyset\) because otherwise \(A_\alpha\) and \(B_\alpha\) would disconnect \(Y_\alpha.\)
Now, by assumption, \(\cap_{\alpha \in T} Y_\alpha\) is non-empty, so pick \(y \in \cap_{\alpha \in T}.\) Then \(y \in A\) or \(y \in B,\) and without loss of generality we can pick \(y \in A.\) Since \(y \in A,\) we have \(y \in A_\alpha\) for all \(\alpha \in T.\) This implies that \(B_\alpha = \emptyset\) for all \(\alpha \in T,\) and in turn \(B = \emptyset.\) This is a contradiction because we assumed \(A\) and \(B\) disconnect \(Y\) and therefore cannot be empty. Therefore \(Y\) is connected.
With this result in place, we are ready to define connected components.
Definition 105 (Connected component)
Let \(X\) be a topological space and \(x \in X.\) Define
Then \(C(x) = \cup_{A \in \mathcal{C}(x)} A\) is the connected component of \(x.\)
In other words, the connected component of \(x \in X\) is the largest connected subset of \(X\) containing \(x,\) which can be shown as follows.
Remark 1 (Connected component of \(x \in X\) is the largest connected subspace containing \(x\))
First, note that \(\{x\} \in \mathcal{C}(x),\) so \(x \in \mathcal{C}(x).\) Second, note that \(C(x)\) is connected because \(x \in \cap_{A \in \mathcal{C}(x)} A\) so the set \(\cap_{A \in \mathcal{C}(x)} A\) is connected, and by our previous lemma, \(C(x)\) must also be connected. Therefore \(C(x)\) is a connected subset of \(X\) that contains \(x.\) Finally, by the defintion of \(C(x),\) we see that if another connected subspace \(C'\) contains \(x,\) then it must be a subset of \(C(x).\)
Similarly to path components, connected components also define an equivalence relation and equivalence classes.
Lemma 32 (Elements in a connected component have the same connected component)
Let \(X\) be a topological space and \(x, y \in X.\) If \(y \in C(x),\) then \(C(x) = C(y).\)
Proof: All elements in a connected component have the same connected component
Let \(X\) be a topological space and \(x, y \in X.\) If \(y \in C(x),\) then \(C(x)\) is a connected subspace containing \(y\) and must therefore be a subset of \(C(y)\) so \(C(x) \subseteq C(y).\) Because of this, we have \(x \in C(y)\) as well, and by the same argument it follows that \(C(y) \subseteq C(x).\) We conclude that \(C(x) = C(y).\)
A corollary of the above is that: the binary relation that says two elements are equivalent if they share the same connected components is an equivalence relation.
Lemma 33 (Connected equivalence relation)
Let \(X\) be a topological space. For \(x, x' \in X,\) define \(x \sim x'\) if \(C(x) = C(x').\) Then \(\sim\) is an equivalence relation.
Proof: Connect equivalence relation
Let \(X\) be a topological space. For \(x, x' \in X,\) define \(x \sim x'\) if \(C(x) = C(x').\) We will prove that this binary relation satisfies the three requirements of equivalence relations. In the following, let \(x, x', x'' \in X.\)
Reflexivity: Trivially, \(C(x) = C(x)\) so \(x \sim x'.\)
Symmetry: Suppose \(x \sim x'.\) This means that \(C(x) = C(x'),\) so \(x' \sim x\) holds as well.
Transitivity: Suppose \(x \sim x'\) and \(x' \sim x''.\) These conditions mean that \(C(x) = C(x')\) and \(C(x') = C(x'')\) so \(C(x) = C(x'')\) and $x \sim x’’.
Lemma 34 (Open connected subspaces of \(\mathbb{R}^n\) are path connected)
If \(U \subseteq \mathbb{R}^n\) with the standard topology is open and connected, then it is path connected.
Proof: open connected subspaces of \(~\mathbb{R}^n\) are path connected
Suppose \(U \subseteq \mathbb{R}^n\) with the standard topology is open and connected. Let \(A\) be a path component of \(U.\) We will show that \(A = U,\) by showing that \(U \setminus A\) must be empty. In turn, we will argue this by showing that both \(A\) and \(U \setminus A\) must be open, and since they are disjoint and \(A\) is non-empty, then \(U \setminus A\) must be empty: otherwise \(A\) and \(U \setminus A\) would disconnect \(U.\)
Showing \(A\) is open: Let \(a \in A.\) Since \(U\) is open, there exists \(\epsilon > 0\) such that \(B_\epsilon(a) \subseteq U.\) Since \(B_\epsilon(a) \simeq \text{Int}(D^n)\) is path connected, and \(A\) is a path component containing \(a,\) we have that \(B_\epsilon(a) \subseteq A.\) Therefore \(A\) is an open subset of \(U.\)
Showing \(U \setminus A\) is open: Now suppose that \(b \in U \setminus A.\) Then since \(U\) is open, there exists \(\epsilon > 0\) such that \(B_\epsilon(b) \subseteq U.\) Now, we must have \(B_\epsilon(b) \cap A = \emptyset\) because if \(B_\epsilon(b)\) and \(A\) intersect, and since \(B_\epsilon(b) \simeq \text{Int}(D^n)\) is path connected, it would follow that \(B_\epsilon(b)\) and \(A\) are path connected, so \(b \in A\) which is a contradiction. Therefore \(B_\epsilon(b) \cap A = \emptyset,\) which means \(B_\epsilon(b) \subseteq U \setminus A,\) which implies that \(U \setminus A\) is open.
We conclude that since \(A\) and \(U \setminus A\) are disjoint open subsets of \(U,\) and \(U\) is connected, we must have \(U \setminus A = \emptyset.\) So \(U = A\) and \(U\) is path connected.