Unit 8: Connectedness, Path-Connectedness & Components

1. Connected Spaces and Separations

Connectedness formalizes the geometric property of a space consisting of a single indivisible piece.

Definition 8.1 (Connected Space & Separation): Let $(X, \mathcal{T})$ be a topological space.

1. A separation of $X$ is a pair of non-empty disjoint open subsets $U, V \subseteq X$ such that:

1. The space $X$ is called connected if there exists no separation of $X$.
2. Equivalently, $X$ is connected if and only if the only clopen subsets (subsets that are simultaneously open and closed) are $\emptyset$ and $X$.

2. Preservation Under Continuous Maps

Theorem 8.1 (Continuous Images of Connected Spaces are Connected): Let $f: X \to Y$ be a continuous map. If $X$ is connected, then the image $f(X)$ is a connected subspace of $Y$.

Proof: Suppose for contradiction that $f(X)$ is not connected. Then there exists a separation of $f(X)$ into two non-empty disjoint open sets in the subspace topology:

Since $f: X \to f(X)$ is continuous, the preimages $f^{-1}(U)$ and $f^{-1}(V)$ are open in $X$.

• Since $U, V \ne \emptyset$, their preimages are non-empty: $f^{-1}(U) \ne \emptyset$ and $f^{-1}(V) \ne \emptyset$.
• $f^{-1}(U) \cap f^{-1}(V) = f^{-1}(U \cap V) = f^{-1}(\emptyset) = \emptyset$.
• $f^{-1}(U) \cup f^{-1}(V) = f^{-1}(U \cup V) = f^{-1}(f(X)) = X$.

Thus $f^{-1}(U)$ and $f^{-1}(V)$ form a separation of $X$! This contradicts the hypothesis that $X$ is connected. Therefore, $f(X)$ must be connected. $\blacksquare$

3. Connected Subsets of $\mathbb{R}$ & The Intermediate Value Theorem

Theorem 8.2 (Connected Subsets of the Real Line): A subset $S \subseteq \mathbb{R}$ is connected if and only if $S$ is an interval (open, closed, half-open, rays, or $\mathbb{R}$).

Theorem 8.3 (Topological Intermediate Value Theorem): Let $X$ be a connected topological space and let $f: X \to \mathbb{R}$ be a continuous function. If $a, b \in X$ and $r \in \mathbb{R}$ is any value between $f(a)$ and $f(b)$ ($f(a) \le r \le f(b)$), then there exists a point $c \in X$ such that:

Proof: By Theorem 8.1, the image $f(X)$ is a connected subset of $\mathbb{R}$. By Theorem 8.2, $f(X)$ is an interval. Since $f(a), f(b) \in f(X)$ and $r$ lies between $f(a)$ and $f(b)$, $r \in f(X)$. Hence there exists $c \in X$ such that $f(c) = r$. $\blacksquare$

1. Connected Components

When a space is not connected, it decomposes uniquely into maximal connected pieces.

Definition 8.2 (Connected Component): Let $X$ be a topological space. For any point $x \in X$, the connected component of $x$, denoted $C(x)$, is the union of all connected subsets of $X$ containing $x$:

Theorem 8.4 (Properties of Connected Components):

1. Each component $C(x)$ is connected.
2. The components of $X$ form a partition of $X$ into mutually disjoint equivalence classes under the relation $x \sim y \iff \exists \text{ connected } S \ni x, y$.
3. Each connected component is a closed subset of $X$.

2. Closures of Connected Sets

Theorem 8.5 (Intermediate Sets are Connected): If $C \subseteq X$ is a connected subset of $X$, and $A \subseteq X$ satisfies:

then $A$ is connected. In particular, the closure $\operatorname{cl}(C)$ of a connected set is always connected.

Proof: Suppose for contradiction that $A$ is not connected. Then there exists a separation of $A$: $A = U \cup V$ with $U, V$ open in $A$, non-empty, and disjoint. Since $C \subseteq A$, $C = (C \cap U) \cup (C \cap V)$. Since $C$ is connected, one of these pieces must be empty, say $C \cap V = \emptyset$. Then $C \subseteq U$. Taking the closure in $A$:

But $V$ is open in $A$ and disjoint from $U$, so $V \cap \operatorname{cl}_A(U) = \emptyset$. Since $V \subseteq A \subseteq \operatorname{cl}_A(U)$, this forces $V = \emptyset$, contradicting $V \ne \emptyset$! Thus $A$ is connected. $\blacksquare$

1. Paths and Path-Connected Spaces

A stronger, more intuitive geometric form of connectedness is the ability to travel between points along a continuous curve.

Definition 8.3 (Path & Path-Connected Space):

1. A path in a topological space $X$ from a point $x$ to a point $y$ is a continuous function:

1. A space $X$ is called path-connected if for every pair of points $x, y \in X$, there exists a path in $X$ connecting $x$ to $y$.

2. Path-Connected Implies Connected

Theorem 8.6: Every path-connected space is connected.

Proof: Suppose $X$ is path-connected. Fix a base point $x_0 \in X$. For every point $x \in X$, there exists a continuous path $\gamma_x: [0, 1] \to X$ with $\gamma_x(0) = x_0$ and $\gamma_x(1) = x$. The image $P_x = \gamma_x([0, 1])$ is the continuous image of the connected interval $[0, 1]$, so $P_x$ is a connected subset of $X$ containing $x_0$. The whole space $X$ is the union of all these paths:

Since each $P_x$ is connected and all share the common point $x_0 \in \bigcap_{x \in X} P_x$, their union $X$ is connected. $\blacksquare$

The converse of Theorem 8.6 is FALSE: a connected space is not necessarily path-connected!

1. Definition of the Topologist's Sine Curve

The classic counterexample separating connectedness from path-connectedness.

Definition 8.4 (Topologist's Sine Curve): In $\mathbb{R}^2$, consider the graph of $\sin(1/x)$ on $(0, 1]$:

The Topologist's Sine Curve is the closure of $S$ in $\mathbb{R}^2$:

is the vertical limit segment on the $y$-axis.

2. Proof that $\bar{S}$ is Connected but Not Path-Connected

Theorem 8.7 (Properties of the Topologist's Sine Curve):

1. $\bar{S}$ is connected.
2. $\bar{S}$ is NOT path-connected.

Proof:

1. Connectedness:

The mapping $x \mapsto (x, \sin(1/x))$ is continuous on the interval $(0, 1]$. Since $(0, 1]$ is connected, its image $S$ is connected. By Theorem 8.5, the closure of any connected set is connected:

1. Failure of Path-Connectedness:

Suppose for contradiction that there exists a continuous path $\gamma: [0, 1] \to \bar{S}$ with:

Let $t_0 = \sup \{t \in [0, 1] : \gamma(t) \in L\}$. Since $L$ is closed, $\gamma(t_0) \in L$. For all $t > t_0$, $\gamma(t) \in S$. Writing $\gamma(t) = (x(t), y(t))$: As $t \to t_0^+$, $x(t) \to 0^+$. Since $y(t) = \sin(1/x(t))$ oscillates infinitely often between $-1$ and $+1$ as $x \to 0^+$, the limit $\lim_{t \to t_0^+} y(t)$ cannot exist! This contradicts the continuity of $\gamma$ at $t_0$. Therefore, no continuous path can bridge $L$ and $S$. Hence $\bar{S}$ is not path-connected. $\blacksquare$

1. Locally Connected and Locally Path-Connected Spaces

Definition 8.5 (Locally Connected & Locally Path-Connected):

1. A topological space $X$ is locally connected if every point has a local basis consisting of connected open sets.
2. A topological space $X$ is locally path-connected if every point has a local basis consisting of path-connected open sets.

Theorem 8.8 (Equivalence for Locally Path-Connected Spaces): If $X$ is locally path-connected, then:

In particular, every topological manifold (which is locally Euclidean) is connected if and only if it is path-connected!

2. Product Invariance of Connectedness

Theorem 8.9 (Products of Connected Spaces are Connected): Let $\{X_\alpha\}_{\alpha \in I}$ be an arbitrary non-empty family of connected topological spaces. Then the product space:

equipped with the product topology is connected.

Proof:

1. For finite products: $X_1 \times X_2$ is covered by horizontal and vertical cross-sections $X_1 \times \{y\}$ and $\{x\} \times X_2$, which are each connected and intersect, proving $X_1 \times X_2$ is connected.
2. For arbitrary products: fix a base point $b = (b_\alpha) \in X$.

For each finite subset $J \subseteq I$, let $X_J = \{x \in X : x_\alpha = b_\alpha \text{ for all } \alpha \notin J\}$. $X_J \cong \prod_{\alpha \in J} X_\alpha$ is connected, and all $X_J$ contain $b$. The union $Y = \bigcup_{J \subseteq I \text{ finite}} X_J$ is connected. By definition of the product topology, $Y$ is dense in $X$: $\bar{Y} = X$. By Theorem 8.5, $X = \bar{Y}$ is connected. $\blacksquare$