Connectedness

Cut a piece of paper cleanly in two and the halves fall apart — no scissors line runs through the middle of either piece. A soap bubble is all one blob; two soap bubbles that have drifted apart are two blobs. Everyone feels the difference between "one piece" and "several pieces". The job of topology is to say exactly what that feeling means using nothing but open sets — no rulers, no coordinates, no straight lines — so that the same word works for a line segment, a space of functions, or the rational numbers.

The surprising part is that the honest definition is phrased in the negative: a space is connected when you cannot tear it cleanly in two. Get comfortable with that double-negative and a whole chain of theorems falls out — culminating in the discovery that the Intermediate Value Theorem you met in first-year calculus is secretly a theorem about connectedness, and that a space can be "all one piece" without your being able to walk from any point to any other. That last twist is the famous topologist's sine curve, and it is waiting for you at the bottom of this page.

The definition: no clean tear

Throughout, X is a topological space (or a subset of one, with the subspace topology). A separation of X is a pair of open sets U, V that split it perfectly:

U \cup V = X, \qquad U \cap V = \varnothing, \qquad U \ne \varnothing, \qquad V \ne \varnothing.

Both pieces are open, they cover everything, they overlap nowhere, and neither is empty. That is a clean tear into two genuine pieces.

X is connected if it admits no separation — you cannot write X = U \cup V with U, V disjoint, nonempty, and both open.
Otherwise X is disconnected, and any such (U, V) is a separation.

Since V = X \setminus U in a separation, each of U and V is the complement of an open set, hence also closed. A set that is both open and closed is called clopen. This gives a second, cleaner phrasing of exactly the same idea.

For a space X, the following are equivalent:

(No separation). X has no separation into two disjoint nonempty open sets.
(Only trivial clopens). The only subsets of X that are both open and closed are \varnothing and X itself.
(No 2-valued map). Every continuous function f : X \to \{0, 1\} (the two-point set with the discrete topology) is constant — there is no continuous surjection onto \{0, 1\}.

Why are these the same statement? If U is a nontrivial clopen set, then (U, X \setminus U) is a separation — and conversely a separation (U, V) makes each piece clopen. And a continuous f : X \to \{0, 1\} is completely determined by the clopen set f^{-1}(0): its preimages of the two (open and closed) points \{0\} and \{1\} are clopen and partition X. So "a nontrivial clopen set", "a separation", and "a continuous surjection onto \{0,1\}" are three names for the same object. Whichever is easiest to rule out in a given problem is the one you use.

A picture of the tear

Below, a single blob on the left has no clean tear — any way you try to cover it with two disjoint open sets, one of them has to be empty. The pair of blobs on the right hands you a separation for free: put one open puff around each, and you have split the whole space into two disjoint nonempty open pieces. Step through the reveal to watch the failed tear on the left and the successful separation on the right.

The canonical examples

An interval in \mathbb{R} is connected. This is the bedrock example — [0,1], (a,b), all of \mathbb{R}. The proof leans on the completeness of the reals (a least-upper-bound argument); we sketch it below. In fact, in \mathbb{R} the connected sets are exactly the intervals (including single points and rays).
The rationals \mathbb{Q} are totally disconnected. Pick any irrational \alpha, say \alpha = \sqrt{2}. Then U = \{\, q \in \mathbb{Q} : q < \sqrt{2} \,\}, \qquad V = \{\, q \in \mathbb{Q} : q > \sqrt{2} \,\} is a separation of \mathbb{Q}: both are open in \mathbb{Q} (each is an open ray intersected with the rationals), they are disjoint, nonempty, and cover \mathbb{Q} because no rational equals \sqrt{2}. The hole where \sqrt{2} should be is exactly the crack the tear runs through. Do this around every irrational and you find the only connected subsets of \mathbb{Q} are single points — it is totally disconnected.
Two disjoint discs are disconnected. In the plane, let X be the union of the open disc of radius 1 around (-2, 0) and the one around (2, 0). Each disc is open, they are disjoint, and together they are X: a textbook separation. Two puddles that never touch are two pieces.
A discrete space with more than one point is disconnected. Every singleton is clopen, so any nonempty proper subset is a nontrivial clopen set. This is the extreme opposite of connected.

Suppose, for contradiction, that [0,1] = U \cup V is a separation, and say 0 \in U. Let

s = \sup \{\, x \in [0,1] : [0, x] \subseteq U \,\}.

This supremum exists because the reals are complete (every bounded set has a least upper bound — the same no-holes property that \mathbb{Q} lacks). Now ask which piece s lands in. If s \in U, then because U is open a little interval around s lies in U, so [0, x] \subseteq U for some x > s — contradicting that s is an upper bound. If s \in V, then because V is open a little interval around s lies in V, so points just below s are in V too, forcing s to be a smaller upper bound. Either way, contradiction. No separation exists, so [0,1] is connected. Every property of intervals you are about to use rests on this one completeness argument.

The theorem that makes it all worthwhile

Connectedness would be a curiosity if it did not travel well under continuous maps. It does — beautifully.

If f : X \to Y is continuous and X is connected, then the image f(X) \subseteq Y is connected.

Proof (one line, using the third face). Suppose f(X) were disconnected. Then there is a continuous surjection g : f(X) \to \{0, 1\}. But then g \circ f : X \to \{0, 1\} is continuous and surjective too (a composition of continuous maps is continuous), contradicting the connectedness of X, which forbids any surjection onto \{0,1\}. Done. The "no 2-valued map" phrasing turned a topological theorem into a two-line algebra of compositions.

Because being connected is preserved by continuous maps, it is a topological invariant: two homeomorphic spaces are either both connected or both disconnected. That is already enough to prove, for instance, that a circle is not homeomorphic to a line segment with its midpoint removed — delete a point from each and count the pieces.

The Intermediate Value Theorem, unmasked

Here is the payoff a calculus student never sees coming. Recall the Intermediate Value Theorem (IVT): if f : [a, b] \to \mathbb{R} is continuous and y lies between f(a) and f(b), then f(c) = y for some c \in [a, b]. The graph cannot jump over the value y; it must pass through it.

Now watch it dissolve into connectedness. The domain [a,b] is connected (previous card). By the image theorem, f([a,b]) is a connected subset of \mathbb{R} — and the connected subsets of \mathbb{R} are exactly the intervals. So the image is an interval containing both f(a) and f(b); an interval has no gaps, so it contains every y between them. Therefore y is attained. That is the IVT.

A continuous real function on a connected domain has a connected image.
In \mathbb{R}, connected = interval.
An interval containing f(a) and f(b) contains everything between them — which is the Intermediate Value Theorem.

The gain is generality. The classical IVT is chained to the real line, but the connectedness version works on any connected domain — a disc, a sphere, a space of functions. "A continuous function cannot skip values on a space that is all one piece" is the real content, and the interval version is just its shadow on \mathbb{R}.

Path-connectedness: the hands-on cousin

The clopen-set definition is powerful but abstract. There is a warmer, more physical notion: a space is path-connected if you can literally walk from any point to any other without leaving it.

X is path-connected if for every pair of points p, q \in X there is a continuous map (a path) \gamma : [0, 1] \to X with \gamma(0) = p and \gamma(1) = q.

Path-connectedness is often the easy thing to check: to show a region of the plane is path-connected, just exhibit a path (a straight segment, if the region is convex; a bent one otherwise). And it implies ordinary connectedness:

Every path-connected space is connected.

Proof sketch. Suppose X were path-connected but disconnected, with a separation (U, V). Pick p \in U and q \in V and a path \gamma between them. Then \big(\gamma^{-1}(U), \gamma^{-1}(V)\big) would be a separation of [0,1] — but [0,1] is connected. Contradiction. The connectedness of the humble interval does all the work.

Warning: the arrow does not reverse. There exist spaces that are connected but not path-connected — you can prove no clean tear exists, yet there is a pair of points with no path between them. The standard witness is the topologist's sine curve, which the "Watch out!" box dissects.

Three tempting-but-wrong instincts trip up nearly everyone meeting connectedness:

"Connected means I can draw a path between any two points." False — that is path-connectedness, a strictly stronger property. The topologist's sine curve is the classic counterexample. Let S = \{\, (x, \sin(1/x)) : 0 < x \le 1 \,\} \;\cup\; \{\, (0, y) : -1 \le y \le 1 \,\}. The curve y = \sin(1/x) oscillates faster and faster as x \to 0^+, swinging between -1 and 1, and we glue on the vertical segment on the y-axis that the oscillations pile up against. This set is connected — the wiggly part has the segment in its closure, and the closure of a connected set is connected, so you cannot separate them. But it is not path-connected: any path trying to reach the segment from the curve would have to survive infinitely many full oscillations in finite time, which no continuous function can do. Connected, yet unwalkable. Scroll down to see it oscillate.
"Connected means convex." No. An annulus (a ring, a washer shape) is connected — it is all one piece — but wildly non-convex: the straight segment between two points can pass through the hole and leave the ring. Convex \Rightarrow path-connected \Rightarrow connected, and none of the arrows reverse.
"I can just look and see it's one piece." Visual intuition is a fine guide for blobs in the plane but a liar in general. The rationals look like they fill the line, yet are totally disconnected; the sine curve looks like two separate strokes of a pen, yet is connected. Trust the open sets, not the eyeball.

Meet the counterexample: y = \sin(1/x)

Watch what happens as x approaches 0 from the right. Each time 1/x advances by 2\pi, the curve completes a full up-and-down swing — and since 1/x \to \infty, those swings come faster and faster, without limit. The graph never settles toward a single height; it keeps sweeping the whole band from -1 to 1 right up to the y-axis. Adding that vertical segment (dashed) gives a set that is one connected piece, yet no path can cross from the wiggles onto the segment.

Connected components: chopping into maximal pieces

A disconnected space still has an honest "how many pieces?" answer. Define a relation on X: say p \sim q if some connected subset of X contains both. This is an equivalence relation, and its classes are the connected components.

The component of a point p is the union of all connected subsets containing p — the largest connected set around p.
The components are connected, pairwise disjoint, and their union is all of X: they partition the space.
Each component is closed. X is connected precisely when it has exactly one component.

Examples: the two disjoint discs have two components (one disc each). The rationals \mathbb{Q} have uncountably many components — every single point is its own component, which is what "totally disconnected" means. The set \mathbb{R} \setminus \{0\} has two components, (-\infty, 0) and (0, \infty); removing a single point cut the line in two, whereas removing a point from the plane \mathbb{R}^2 \setminus \{0\} leaves it connected. Counting components after a puncture is one of the sharpest cheap tools for telling spaces apart.

Components count pieces, not holes. A coffee cup (with its handle) is a single connected piece — you can walk from the rim to the base of the handle without leaving the surface — so it has exactly one component, just like a solid ball or a doughnut. Connectedness cannot tell a doughnut from a ball; that job belongs to finer invariants (the fundamental group, homology) which count holes and loops. Think of connectedness as topology's coarsest question: "how many separate pieces?" It is the first thing you ask, the cheapest to answer, and the sanity check every other invariant is built on top of.

Worked example: how many components has \{1/n\} \cup \{0\}?

Let X = \{\, 1/n : n = 1, 2, 3, \dots \,\} \cup \{0\}, viewed inside \mathbb{R}. How many connected components does it have?

Each point 1/n is isolated. Around 1/n there is a tiny interval containing no other point of X, so \{1/n\} is open in X — and also closed. It is a clopen singleton, hence its own component. There are countably infinitely many such points.

The point 0 is different but still alone. No neighbourhood of 0 is a single point (every interval around 0 catches infinitely many 1/n), yet \{0\} is still a component: the only connected subsets of this space are single points, because between any two distinct points of X you can slide an irrational cut and separate them.

Answer: countably infinitely many components, one per point. The space is totally disconnected — even though, drawn on the line, its points visibly bunch up toward 0. Accumulation is not connection: the points crowd together but never touch, so no two of them share a component.