Connectedness · study.

An interval on the real line has a feature no union of separated pieces has: one can move continuously from any point to any other without leaving it or jumping a gap. Connectedness abstracts exactly that. Like compactness, it is a topological invariant preserved by continuous maps; compactness underlies the extreme value theorem, and connectedness underlies the intermediate value theorem: a continuous function cannot skip a value on a connected domain because its image is again connected.

Separations

The definition rules out the way a space could split into two pieces.

Equivalently, $X$ is disconnected when it splits as $X = X_{1} \cup X_{2}$ with $X_{1}, X_{2}$ nonempty, open, and disjoint; such a pair is a separation. Each piece is then also closed (it is the complement of the other), so a separation is precisely a nontrivial clopen decomposition. Connectedness is the absence of any separation: the space is all one piece.

For a subset $S$ of a larger space, the separating sets are taken in the ambient space and cut down to $S$ .

The sets $U_{1}, U_{2}$ need not be disjoint in $X$ ; they only have to separate the points of $S$ , meeting no common point of $S$ . This subtlety is why the condition is phrased through $U_{1} \cap U_{2} \cap S = \emptyset$ rather than $U_{1} \cap U_{2} = \emptyset$ .

A connected set (left) has no separation; a disconnected set (right) splits into two nonempty open pieces with a gap between them.

A ball need not be connected either. In the two-point discrete space ${a, b}$ , the ball $B (a, 2) = {a, b}$ splits into the open singletons ${a}$ and ${b}$ .

Connected subsets of the line

On $R$ connectedness has a complete, concrete description.

Proof. Suppose $S$ is connected and not a single point, and take $x, y \in S$ with $x < y$ . If some $z$ with $x < z < y$ were missing from $S$ , the open rays $(- \infty, z)$ and $(z, \infty)$ would meet $S$ (at $x$ and $y$ ), be disjoint, and jointly cover $S$ — a separation. Connectedness forbids this, so $z \in S$ : the set contains every point between any two of its points, which is just the definition of an interval.

Conversely, let $S$ be an interval and suppose open sets $U_{1}, U_{2}$ satisfy $U_{1} \cap S \neq = \emptyset$ , $U_{2} \cap S \neq = \emptyset$ , and $S = (U_{1} \cap S) \cup (U_{2} \cap S)$ . We find a common point, so the pieces are not disjoint on $S$ and no separation exists. Take $x \in U_{1} \cap S$ , $y \in U_{2} \cap S$ , without loss of generality $x < y$ ; as $S$ is an interval, $[x, y] \subseteq S$ . The set $U_{2} \cap [x, y]$ is nonempty, so the least-upper-bound property supplies

z = inf (U_{2} \cap [x, y]) .

If $z = x$ , then $z \in U_{1}$ . If $z > x$ , then every ball $(z - ε, z + ε)$ contains points of $[x, y]$ below $z$ , hence outside $U_{2}$ ; an open set containing $z$ would trap such points, so $z \in / U_{2}$ , and since every point of $[x, y]$ lies in $U_{1}$ or $U_{2}$ , $z \in U_{1}$ . Now $U_{1}$ is open, so $B (z, δ) \subseteq U_{1}$ for some $δ > 0$ ; and $z$ being the infimum of $U_{2} \cap [x, y]$ , some $w \in U_{2} \cap [x, y]$ lies in $[z, z + δ) \subseteq U_{1}$ . Then $w \in U_{1} \cap U_{2} \cap S$ . $■$

A subset of

R

with a hole:

x

and

y

are in

S

but the point

z

between them is not, so the two rays separate

S

This underlies the intermediate value theorem: intervals are connected, continuous maps preserve connectedness, so the image of an interval is again an interval and omits no value between two it attains.

Path-connectedness

Verifying that no separation exists is awkward; exhibiting an actual path is constructive and usually easier.

If a path-connected $A$ had a separation $A = (U_{1} \cap A) \cup (U_{2} \cap A)$ , take $p \in U_{1}$ , $q \in U_{2}$ , and a path $γ$ between them; then $γ^{- 1} (U_{1})$ and $γ^{- 1} (U_{2})$ would separate $[0, 1]$ , contradicting its connectedness. So no separation exists.

Path-connectedness: any two points

p

and

q

of the region are joined by a continuous path

γ

that stays inside it.

The implication does not reverse in general. The topologist's sine curve, the closure of ${(x, sin (1/ x)) : 0 < x \leq 1}$ together with the segment ${0} \times [- 1, 1]$ , is connected: the oscillating graph accumulates on the whole vertical segment, so no separation can split the segment from the graph. Yet no continuous path reaches the segment from the graph, because a path would have to traverse infinitely many oscillations of unit height in finite time. So the curve is connected but not path-connected.

The topologist's sine curve: the graph of

sin (1/ x)

oscillates ever faster near

0

and accumulates on the vertical segment, giving a connected set that no path can cross from graph to segment.

For open subsets of $R^{n}$ , however, the two notions coincide, so on regions in Euclidean space one may use whichever is convenient. Convex sets and star-shaped sets are path-connected outright: the straight segment $γ (t) = (1 - t) p + tq$ stays inside them, so balls, boxes, and all of $R^{n}$ are connected.

A product of connected sets is connected: $[0, 1] \times [0, 1]$ is connected because any two of its points are joined by an L-shaped path (horizontal then vertical) lying inside the square. The same argument makes any box in $R^{n}$ connected, and hence, by the theorem below, the target of an intermediate value statement in several variables.

The unit square is path-connected: an L-shaped path (horizontal, then vertical) joins any two points without leaving the square.

Components and invariance

Every space partitions into maximal connected pieces.

Components are always closed; in $Q$ they are single points, and in an interval there is one component, the whole interval.

The two components of

[0, 1] \cup [2, 3]

: disjoint open sets separate the pieces, and each piece is a maximal connected subset.

The number and structure of components is preserved under any map that preserves connectedness, and continuous maps do.

A separation of $f (X)$ by open $V_{1}, V_{2} \subseteq Y$ would pull back, via the preimage characterization of continuity, to a separation of $X$ by the open sets $f^{- 1} (V_{1})$ and $f^{- 1} (V_{2})$ , contradicting connectedness of $X$ . So connectedness is a topological invariant: it is preserved by continuous maps and, in particular, by homeomorphisms.

The intermediate value theorem is the immediate corollary in one variable.

The interval $[a, b]$ is connected, so its image $f ([a, b])$ is a connected subset of $R$ , hence an interval; an interval containing $f (a)$ and $f (b)$ contains every value between them, so $y$ is attained. No estimate on $f$ is used, only that a continuous image of a connected set is connected. Connectedness supplies the no gaps in the domain half of the theorem the way compactness supplies the attains its bounds half of the extreme value theorem.

Property	Definition via	Preserved by continuous maps	Controls
Compactness	finite subcovers	image is compact	extreme value theorem
Connectedness	no separation	image is connected	intermediate value theorem

Both invariance statements reappear for continuous functions on metric spaces: the image of a compact set is compact gives the extreme value theorem, and the image of a connected set is connected gives the intermediate value theorem.¹

Lebl, Basic Analysis I, §7.2.2 — Connected sets: clopen decompositions and separations, the characterization of connected subsets of $R$ as intervals, and connectedness under the subspace topology. ↩