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 $f:X \to \{0,1\}.$

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 U, V 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. ♦

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: (a, b) is connected.

Since (a, b) 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 a < r < b 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>a, x is not a lower bound of X so there exists y<x, y 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 [a, b] is a compact and connected subset of R, so is f([a, b]). Out of the 9 possible connected subsets of R, only the closed interval [c, d] is closed and bounded. Thus, f([a, b]) = [c, d]. ♦

Connected Components

Let X be a topological space. For any two points x, y in X, write x ~ y 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.

We 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 X = Q (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 r, a<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 X_j 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. ♦

The 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. ]

Examples

An infinite set with the cofinite topology is connected since all non-empty closed subsets are finite and hence not open.
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.
By proposition 6, the square [0, 1] × [0, 1] is connected, as is its interior (0, 1) × (0, 1) in R².
The circle S¹ is connected since it’s the continuous image of $t\mapsto (\cos(t),\sin(t)).$
(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.

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 X = R and Y be the union of (-1, 0) and (0, 1); cl(Y) = [-1, 1]. Second statement: let X = R² 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. ♦