Topology: Connected Spaces

Let X be a topological space. Recall that if U is a clopen (i.e. open and closed) subset of X, then X is the topological disjoint union of U and X-U. Hence, if we assume X cannot be decomposed any further, there’re no non-trivial clopen subsets of X.

Definition. The space X is connected if its only clopen subsets are X and the empty set. Equivalently, there’re no non-empty disjoint open subsets U, V of X such that X = U\cup V. Otherwise, it’s disconnected.

The following characterisation of connected sets is simple but surprisingly handy.

Theorem 1. X is connected if and only if any continuous map f:X \to \{0, 1\}, where {0, 1} has the discrete topology, is constant.

Thus, it’s disconnected if and only if there’s a continuous surjective map f:X \to \{0,1\}.

disconnected

Proof.

If f is not constant, then U := f^{-1}(0) and V:=f^{-1}(1) are non-empty disjoint open subsets such that X=U\cup V. Conversely, if X = U\cup V, where UV are non-empty, disjoint and open in X, then we can set

f:X\to \{0, 1\}, \quad x\mapsto \begin{cases} 0, &\quad\text{if } x\in U, \\ 1, &\quad\text{if } x\in V.\end{cases} ♦

With this theorem in hand, the remaining properties are surprisingly easy to prove.

Proposition 2. If Y is a connected subset of X, then its closure Z := \text{cl}_X(Y) is connected.

Proof.

First \text{cl}_Z(Y) = Z\cap\text{cl}_X(Y) = Z, so Y is dense in Z. Suppose f:Z\to\{0, 1\} is continuous. Then f|_Y is constant (say, 0) since Y is connected. Since f^{-1}(0) is closed and contains Y, it must be the whole of Z. ♦

Proposition 3. If f:X\to Y is a continuous map of topological spaces and X is connected, then so is f(X).

Proof.

Replace Y by f(X); we may assume f is surjective. Suppose g:Y\to \{0, 1\} is a surjective continuous map. Then g\circ f : X\to \{0, 1\} is also continuous and surjective, so X is disconnected (contradiction). Hence, g does not exist. ♦

The union of connected subsets is not connected in general; in fact, topological disjoint unions of two or more non-empty spaces are never connected. But if the spaces share a common point, we have the following.

Proposition 4. Let \{Y_i\} be a collection of connected subspaces of X. If \cap_i Y_i \ne \emptyset, then Y:= \cup_i Y_i is connected.

Proof.

Pick x\in \cap_i Y_i. Let f:Y\to\{0, 1\} be a continuous map. Then each f|_{Y_i} is constant for each i, and it must be equal to f(x). So f is constant. ♦

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Intermediate Value Theorem

Let’s consider subsets of R. We claim:

Lemma. A subset of R is connected if and only if it’s one of the following:

  • bounded: (a, b), (a, b], [a, b), [a, b];
  • unbounded: (-\infty, b), (-\infty, b], (a, \infty), [a, \infty), \mathbf{R};

Proof.

The proof is long but conceptually easy. First we show that (a, b) is connected.

  • If not, write (a, b) as a disjoint union of non-empty open subsets U and V. Let x = sup(U) ≤ b.
  • Now x doesn’t lie in U. Otherwise x<b, so N(x, 2ε) lies in U for some ε>0. In particular, x+ε lies in U, so x is not an upper bound of U (contradiction).
  • So x lies in V. But that means for some ε>0, N(x, ε) lies in V and hence is disjoint from U. That means x-ε is an upper bound of U (contradiction).
  • Conclusion: (ab) is connected.

Since (ab) is dense in all the bounded intervals (a, b], [a, b), [a, b], these are all connected by proposition 2. Finally, for the unbounded intervals, apply proposition 4. E.g. (0, \infty) = \cup_{n=1}^\infty (0, n). 

To prove the converse, we note that if X\subseteq \mathbf{R} is connected and a,b\in X with a<b, then [a,b]\subseteq X. [ Indeed, if arb and r\not\in X, then X = ((-\infty, r)\cap X)\cup (r,\infty) \cap X) and the two open subsets are non-empty since they contain a and b respectively. ]

Now let b = sup(X). There’re 3 possibilities: either (i) b = ∞, (ii) b < ∞ and lies in X, or (iii) b < ∞ and doesn’t lie in X. The same holds for a = inf(X): either (a) a = -∞, (b) a > -∞ and lies in X, or (c) a > -∞ and doesn’t lie in X. Each of the 9 possibilities gives rise to an interval above. We’ll leave the details to the reader.

[ Example for case (i)(b). For each x>ax is not a lower bound of X so there exists y<xy lies in X. Since sup(X) = ∞, there exists z>x which lies in X. So [y,z]\subseteq X and x\in X. Thus, (a,\infty) \subseteq X. The fact that a isn't in X shows that equality holds. ] ♦

Corollary (Intermediate Value Theorem). If f:[a,b]\to \mathbf{R} is continuous, then the image of f is also a closed interval [c, d]. In particular, if s lies between f(a) and f(b), then there exists an r in [a, b], f(r) = s.

Proof.

Since [ab] is a compact and connected subset of R, so is f([ab]). Out of the 9 possible connected subsets of R, only the closed interval [cd] is closed and bounded. Thus, f([ab]) = [cd]. ♦

blue-linConnected Components

Let X be a topological space. For any two points xy in X, write xy if there’s a connected subset of X containing them. We claim this gives an equivalence relation; indeed, it is clearly reflexive and symmetry. Transitivity follows straight from proposition 3.

Definition. The equivalence classes of this relation are called the connected components of X.

Note that each connected component Y is a maximal connected subset of X, in the sense that if we add any point outside Y, the resulting set won’t be connected.

warningWe know that X is a set-theoretic disjoint union of its connected components. It’s tempting to think that X is a topological disjoint union as well. But that’s wrong. Indeed, let’s look at XQ (as a subspace of R). The only connected subsets of X are the singleton points {x}: for if Y\subseteq X and a,b\in Y with a<b, then we can pick an irrational ra<r<b so that Y = ((-\infty, r)\cap Y) \cup ((r, +\infty)\cap Y) is a disjoint union of two non-empty open subsets. So Y is disconnected.

Thus, the connected components of Q are the singleton sets. But Q is clearly not discrete.

Definition. A space whose connected components are all singleton sets is said to be totally disconnected.

Thus, a discrete space is totally disconnected but not vice versa.

The example of Q shows that connected components are in general not open. However, they’re closed.

Proposition 5. Every connected component Y of a topological space X is closed.

Proof.

This follows almost immediately from proposition 2: since Y is connected, so is \text{cl}_X(Y)\supseteq Y. But Y is a maximal connected subset, so Y = \text{cl}_X(Y). ♦

Proposition 6. If \{X_i\} is a collection of non-empty topological spaces, the product X := \prod_i X_i is connected if and only if each X_i is connected.

Proof.

(→) is obvious since each X_i is the continuous image of the projection map from X. For (←), let f:X\to \{0,1\} be surjective and continuous. Let (x_i)\in f^{-1}(0). Since f^{-1}(0) is open, it contains a basic open set of the form \prod_i U_i where U_i=X_i for all but finitely many i‘s : \{i_1, \ldots, i_n\}. Thus:

\left(\prod_{i\ne i_1,\ldots, i_n} X_i\right) \times \{x_{i_1}\} \times \{x_{i_2}\} \times\ldots\times \{x_{i_n}\} \subseteq f^{-1}(0). (#)

Next, use the following.

  • Sublemma. If j is an index, and (x_i), (y_i) \in X satisfy: x_i = y_i for all i except i=j, then they belong to the same connected component.
  • Proof. The subspace X_j \times (\prod_{i\ne j} \{x_i\}) is homeomorphic to Xj so it is a connected subset containing (x_i), (y_i).

Hence on the LHS of (#), we can change the coordinates x_{i_1}, \ldots, x_{i_n} one at a time and see that the whole \prod_i X_i is contained in f^{-1}(0), i.e. X is connected. ♦

warningThe above sublemma holds for the box topology too. This may con us into believing that \prod_i X_i under the box topology is also connected. But let’s take X=\mathbf{R}^\mathbf{N}, the space of all real sequences and U the subset of all sequences converging to 0. Any (x_n) \in U is also contained in \prod_{n=1}^\infty (x_n - \frac 1 n, x_n + \frac 1 n) which is open in the box topology and contained in U. Thus U is open. The same holds for any (x_n)\in X-U so X-U is also open. [ Note that the sublemma implies altering a single term in a sequence has no effect on whether it converges to 0. ]

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Examples

  1. An infinite set with the cofinite topology is connected since all non-empty closed subsets are finite and hence not open.
  2. The space X = [0, 1)\cup (2, 3] is disconnected since [0, 1) = (-1, 1) \cap X and (2, 3] = (2, 4)\cap X are open subsets of X. These give the connected components of X.
  3. By proposition 6, the square [0, 1] × [0, 1] is connected, as is its interior (0, 1) × (0, 1) in R2.
  4. The circle S1 is connected since it’s the continuous image of t\mapsto (\cos(t),\sin(t)).
  5. (Topologist’s Sine Curve) Take the set X = Y\cup Z \subset \mathbf{R}^2, where Y = \{0\}\times [0, 1], Z=\{(x, \sin(1/x)) : 0 < x \le 1\}. Now Y and Z are homeomorphic to [0, 1] and (0, 1] respectively, so they’re connected. What’s surprising is that X is connected as well! Indeed, one observes that Y is a set of accumulation points for Z, since for every open point (0, y) in Y and ε>0, N_X((0, y), \epsilon) contains a point of Z. Thus \text{cl}_X(Z) = X. Since Z is connected, proposition 2 tells us X is connected too.

topo_sine_curve

Exercise.

Which of the following is/are correct for a subset Y of X?

  • If cl(Y) is connected, then so is Y.
  • If Y is connected, then so is int(Y).
  • If int(Y) is connected, then so is Y.

Answer (Highlight to Read).

They’re all wrong! First statement: let XR and Y be the union of (-1, 0) and (0, 1); cl(Y) = [-1, 1]. Second statement: let XR2 and Y be the union of [-1, 0] × [-1, 0] and [0, 1] × [0, 1]. Each square is connected; since they share a common point, Y is connected. But int(Y) is the disjoint union of two open squares. Third statement: let Y be {0, 1} in R; int(Y) is empty and hence connected. ♦

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