Connectedness in R

A set is connected if it cannot be split into two disjoint nonempty open pieces. In $\mathbb{R}$ , the connected sets are precisely the intervals (including singletons, rays, and the whole line). Connectedness is the topological property underlying the Intermediate Value Theorem: continuous images of connected sets are connected.

Definition

Definition3.1Connected set

A set $E \subseteq \mathbb{R}$ is connected if it is not the union of two disjoint nonempty open sets (in the subspace topology). That is, $E$ is connected if there do not exist open sets $U, V \subseteq \mathbb{R}$ such that:

$U \cap E \neq \varnothing$ and $V \cap E \neq \varnothing$ ,
$(U \cap E) \cap (V \cap E) = \varnothing$ ,
$E \subseteq U \cup V$ .

A set that is not connected is disconnected.

RemarkIntuition

Connectedness means the set has "no gaps" — you cannot separate it into two pieces with a gap in between. In $\mathbb{R}$ , this is equivalent to being an interval.

ExampleIntervals are connected

Any interval $I$ (open, closed, half-open, bounded or unbounded) is connected. For instance, $[a, b]$ , $(a, b)$ , $[a, \infty)$ , and $\mathbb{R}$ are all connected. The proof uses the completeness of $\mathbb{R}$ : if $I = U \cup V$ with $U, V$ disjoint open, then choosing $a \in U$ and $b \in V$ , the supremum $\sup(U \cap [a, b])$ must lie in either $U$ or $V$ , leading to a contradiction.

ExampleDisconnected sets

The set $E = [0, 1] \cup [2, 3]$ is disconnected. Take $U = (-1, 1.5)$ and $V = (1.5, 4)$ . Then $U \cap E = [0, 1]$ and $V \cap E = [2, 3]$ are disjoint nonempty open sets in $E$ , and $E = (U \cap E) \cup (V \cap E)$ .

ExampleQ is disconnected

The rationals $\mathbb{Q}$ are disconnected. For any irrational $\alpha$ (e.g., $\sqrt{2}$ ), let $U = (-\infty, \alpha)$ and $V = (\alpha, \infty)$ . Then $U \cap \mathbb{Q}$ and $V \cap \mathbb{Q}$ are nonempty, disjoint, and cover $\mathbb{Q}$ . So $\mathbb{Q}$ is not connected.

Characterization of connected sets in R

Theorem3.1Connected sets are intervals

A subset $E \subseteq \mathbb{R}$ is connected if and only if $E$ is an interval. That is, $E$ is connected iff whenever $x, y \in E$ with $x < y$ , every $z \in (x, y)$ also lies in $E$ .

RemarkProof sketch

( $\Leftarrow$ ) If $E$ is an interval and $E = U \cup V$ with $U, V$ disjoint open in $E$ , choose $a \in U$ , $b \in V$ with $a < b$ . Let $c = \sup\{t \in [a, b] \mid [a, t] \subseteq U\}$ . Then $c \in E$ , so $c \in U$ or $c \in V$ . Both cases lead to contradictions using the openness of $U$ and $V$ .

( $\Rightarrow$ ) If $E$ is not an interval, there exist $x, y \in E$ with $x < y$ and some $z \in (x, y)$ with $z \notin E$ . Then $U = (-\infty, z)$ and $V = (z, \infty)$ separate $E$ .

ExampleAll types of intervals are connected

The following are all connected:

Bounded closed: $[a, b]$ .
Bounded open: $(a, b)$ .
Half-open: $[a, b)$ , $(a, b]$ .
Rays: $[a, \infty)$ , $(a, \infty)$ , $(-\infty, b]$ , $(-\infty, b)$ .
The whole line: $\mathbb{R}$ .
Singletons: $\{a\}$ .

Intermediate Value Theorem

Theorem3.2Continuous images of connected sets

If $E$ is connected and $f : E \to \mathbb{R}$ is continuous, then $f(E)$ is connected.

RemarkIntermediate Value Theorem

Since $E$ connected in $\mathbb{R}$ means $E$ is an interval, and $f(E)$ connected means $f(E)$ is an interval, Theorem 3.2 implies the Intermediate Value Theorem (IVT):

If $f : [a, b] \to \mathbb{R}$ is continuous, then $f$ takes on every value between $f(a)$ and $f(b)$ .

ProofIVT from connectedness

If $f(a) < c < f(b)$ , then $c \in [f(a), f(b)]$ . Since $[a, b]$ is connected, $f([a, b])$ is connected, hence an interval. Thus $[f(a), f(b)] \subseteq f([a, b])$ , so there exists $x \in [a, b]$ with $f(x) = c$ .

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ExampleExistence of roots

Let $f(x) = x^3 - x - 1$ . Then $f(1) = -1 < 0$ and $f(2) = 5 > 0$ . By the IVT, there exists $c \in (1, 2)$ with $f(c) = 0$ . Thus the cubic has a real root between $1$ and $2$ .

ExampleFixed point theorem

If $f : [0, 1] \to [0, 1]$ is continuous, then $f$ has a fixed point: there exists $x \in [0, 1]$ with $f(x) = x$ . Proof: consider $g(x) = f(x) - x$ . Then $g(0) = f(0) \geq 0$ and $g(1) = f(1) - 1 \leq 0$ . By IVT, $g(c) = 0$ for some $c \in [0, 1]$ .

Path-connectedness

Definition3.2Path-connected

A set $E \subseteq \mathbb{R}$ is path-connected if for any two points $x, y \in E$ , there exists a continuous function $\gamma : [0, 1] \to E$ (a path) with $\gamma(0) = x$ and $\gamma(1) = y$ .

RemarkIn R, connected = path-connected

For subsets of $\mathbb{R}$ , connectedness and path-connectedness are equivalent. Both are equivalent to being an interval. In higher dimensions (e.g., $\mathbb{R}^2$ ), there are connected sets that are not path-connected (e.g., the topologist's sine curve).

Summary

Connectedness characterizes intervals in $\mathbb{R}$ :

Connected sets are intervals (no gaps).
Continuous images of connected sets are connected (IVT).
Path-connectedness is equivalent to connectedness in $\mathbb{R}$ .

See Intermediate Value Theorem for detailed applications, and Metric Spaces for generalizations.