Connected Subsets of the Real Line

The characterization of connected subsets of $\mathbb{R}$ as precisely the intervals is one of the most fundamental results in point-set topology. It connects the order structure, the completeness property, and the topology of the real line.

Main Theorem

Theorem4.13Connected Subsets of $\mathbb{R}$

A subset $A \subseteq \mathbb{R}$ is connected (in the standard topology) if and only if $A$ is an interval. Here "interval" includes:

$\emptyset$ , singletons $\{a\}$ , and $\mathbb{R}$ ,
bounded intervals $(a, b)$ , $[a, b]$ , $[a, b)$ , $(a, b]$ ,
unbounded intervals $(-\infty, b)$ , $(-\infty, b]$ , $(a, \infty)$ , $[a, \infty)$ .

Equivalently, $A$ is an interval if and only if for all $a, b \in A$ with $a < b$ , every $c$ with $a < c < b$ satisfies $c \in A$ .

Proof

( $\Rightarrow$ : Connected implies interval.) We prove the contrapositive. Suppose $A$ is not an interval. Then there exist $a, b \in A$ and $c \notin A$ with $a < c < b$ . Define: $U = A \cap (-\infty, c), \qquad V = A \cap (c, \infty).$

Since $c \notin A$ , we have $A = U \cup V$ . Both $U$ and $V$ are open in $A$ (they are intersections of $A$ with open sets in $\mathbb{R}$ ). They are nonempty ( $a \in U$ , $b \in V$ ) and disjoint. So $\{U, V\}$ is a separation of $A$ , and $A$ is disconnected.

( $\Leftarrow$ : Interval implies connected.) Let $I$ be an interval. Suppose for contradiction that $I = U \cup V$ is a separation. Choose $a \in U$ and $b \in V$ ; without loss of generality, $a < b$ .

Since $I$ is an interval, $[a, b] \subseteq I$ . Define: $s = \sup(U \cap [a, b]).$

This supremum exists since $U \cap [a, b]$ is nonempty (contains $a$ ) and bounded above by $b$ .

Claim: $s \notin U$ .

Suppose $s \in U$ . Since $U$ is open in $I$ , there exists $\epsilon > 0$ with $(s - \epsilon, s + \epsilon) \cap I \subseteq U$ . If $s < b$ , then for $\delta = \min(\epsilon, b - s) / 2$ , the point $s + \delta \in [a, b]$ and $s + \delta \in U$ , contradicting $s = \sup(U \cap [a, b])$ . If $s = b$ , then $b \in U \cap V = \emptyset$ , a contradiction. So $s \notin U$ .

Claim: $s \notin V$ .

Since $s \notin U$ , we must have $s \in V$ (as $s \in [a, b] \subseteq I = U \cup V$ ). Since $V$ is open in $I$ , there exists $\epsilon > 0$ with $(s - \epsilon, s + \epsilon) \cap I \subseteq V$ . In particular, $(s - \epsilon, s) \cap I \subseteq V$ , so $(s - \epsilon, s) \cap U \cap [a, b] = \emptyset$ . This means $U \cap [a, b] \subseteq [a, s - \epsilon]$ , contradicting $s = \sup(U \cap [a,b])$ (since $s > a$ and $\epsilon > 0$ ).

Since $s \notin U$ and $s \notin V$ but $s \in I = U \cup V$ , we have a contradiction. Therefore $I$ is connected.

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Consequences

Theorem4.14Continuous Functions on Intervals

Let $f: I \to \mathbb{R}$ be continuous where $I$ is an interval. Then $f(I)$ is an interval.

Proof

$I$ is connected (by Theorem 4.13). The continuous image of a connected space is connected. A connected subset of $\mathbb{R}$ is an interval (by Theorem 4.13).

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ExampleNo Continuous Bijection $\mathbb{R} \to \mathbb{R}^2$

There is no continuous bijection $f: \mathbb{R} \to \mathbb{R}^2$ . Indeed, removing a point $p$ from $\mathbb{R}$ disconnects it (giving two components), while removing $f(p)$ from $\mathbb{R}^2$ leaves a connected space ( $\mathbb{R}^2 \setminus \{q\}$ is path-connected for any $q$ ). Since $f$ restricted to $\mathbb{R} \setminus \{p\} \to \mathbb{R}^2 \setminus \{f(p)\}$ is continuous and injective, the image would be a connected subset mapping bijectively from a disconnected domain -- this leads to a contradiction via more careful analysis.

More rigorously, this argument shows $\mathbb{R} \not\cong \mathbb{R}^2$ , and the general invariance of domain theorem shows $\mathbb{R}^m \not\cong \mathbb{R}^n$ for $m \neq n$ .

Extensions to Ordered Spaces

Theorem4.15Connected Ordered Spaces

A totally ordered set $L$ with the order topology is connected if and only if:

$L$ has the least upper bound property: every nonempty subset bounded above has a supremum.
$L$ is dense: for every $a < b$ in $L$ , there exists $c \in L$ with $a < c < b$ .

ExampleConnected and Disconnected Ordered Spaces

$\mathbb{R}$ satisfies both conditions: it is connected.
$\mathbb{Q}$ is dense but lacks the LUB property: it is disconnected.
$\mathbb{Z}$ has the LUB property (vacuously for bounded sets) but is not dense: it is disconnected (discrete).
The long line $L$ satisfies both conditions: it is connected.
$[0, 1] \times [0, 1]$ in dictionary order satisfies both conditions: it is connected. The induced topology differs from the product topology (it is the "ordered square").

RemarkCharacterization of $\mathbb{R}$

Up to order isomorphism, $\mathbb{R}$ is the unique connected, separable, linearly ordered set without endpoints. This characterization uses the LUB property (for connectedness), a countable dense subset (for separability), and the existence of elements above and below any given element (no endpoints).