Let *X* be a topological space. Recall that if *U* is a clopen (i.e. open and closed) subset of *X*, then *X* is the topological disjoint union of *U* and *X*-*U*. Hence, if we assume *X* cannot be decomposed any further, there’re no non-trivial clopen subsets of *X*.

Definition. The space X isconnectedif its only clopen subsets are X and the empty set. Equivalently, there’re no non-empty disjoint open subsets U, V of X such that Otherwise, it’sdisconnected.

The following characterisation of connected sets is simple but surprisingly handy.

Theorem 1. X is connected if and only if any continuous map where {0, 1} has the discrete topology, is constant.Thus, it’s disconnected if and only if there’s a continuous surjective map

**Proof**.

If *f* is not constant, then and are non-empty disjoint open subsets such that Conversely, if where *U*, *V* are non-empty, disjoint and open in *X*, then we can set

♦

With this theorem in hand, the remaining properties are surprisingly easy to prove.

Proposition 2. If Y is a connected subset of X, then its closure is connected.

**Proof**.

First so *Y* is dense in *Z*. Suppose is continuous. Then is constant (say, 0) since *Y* is connected. Since is closed and contains *Y*, it must be the whole of *Z*. ♦

Proposition 3. If is a continuous map of topological spaces and X is connected, then so is f(X).

**Proof**.

Replace *Y* by *f*(*X*); we may assume *f* is surjective. Suppose is a surjective continuous map. Then is also continuous and surjective, so *X* is disconnected (contradiction). Hence, *g* does not exist. ♦

The union of connected subsets is not connected in general; in fact, topological disjoint unions of two or more non-empty spaces are never connected. But if the spaces share a common point, we have the following.

Proposition 4. Let be a collection of connected subspaces of X. If then is connected.

**Proof**.

Pick . Let be a continuous map. Then each is constant for each *i*, and it must be equal to *f*(*x*). So *f* is constant. ♦

## Intermediate Value Theorem

Let’s consider subsets of **R**. We claim:

Lemma. A subset ofRis connected if and only if it’s one of the following:

- bounded: ;
- unbounded: ;

**Proof**.

The proof is long but conceptually easy. First we show that (*a*, *b*) is connected.

- If not, write (
*a*,*b*) as a disjoint union of*non-empty*open subsets*U*and*V*. Let*x*= sup(*U*) ≤*b*. - Now
*x*doesn’t lie in*U*. Otherwise*x*<*b*, so*N*(*x*, 2ε) lies in*U*for some ε>0. In particular,*x*+ε lies in*U*, so*x*is not an upper bound of*U*(contradiction). - So
*x*lies in*V*. But that means for some ε>0,*N*(*x*, ε) lies in*V*and hence is disjoint from*U*. That means*x*-ε is an upper bound of*U*(contradiction). - Conclusion: (
*a*,*b*) is connected.

Since (*a*, *b*) is dense in all the bounded intervals these are all connected by proposition 2. Finally, for the unbounded intervals, apply proposition 4. E.g.

To prove the converse, we note that if is connected and with *a*<*b*, then [ Indeed, if *a* < *r* < *b* and then and the two open subsets are non-empty since they contain *a* and *b* respectively. ]

Now let *b* = sup(*X*). There’re 3 possibilities: either (i) *b* = ∞, (ii) *b* < ∞ and lies in *X*, or (iii) *b* < ∞ and doesn’t lie in *X*. The same holds for *a* = inf(*X*): either (a) *a* = -∞, (b) *a* > -∞ and lies in *X*, or (c) *a* > -∞ and doesn’t lie in *X*. Each of the 9 possibilities gives rise to an interval above. We’ll leave the details to the reader.

[ Example for case (i)(b). For each *x*>*a*, *x* is not a lower bound of *X* so there exists *y*<*x*, *y* lies in *X*. Since sup(*X*) = ∞, there exists *z*>*x* which lies in *X*. So Thus, The fact that *a* isn’t in *X* shows that equality holds. ] ♦

Corollary (Intermediate Value Theorem). If is continuous, then the image of f is also a closed interval [c, d]. In particular, if s lies between f(a) and f(b), then there exists an r in [a, b], f(r) = s.

**Proof**.

Since [*a*, *b*] is a compact and connected subset of **R**, so is *f*([*a*, *b*]). Out of the 9 possible connected subsets of **R**, only the closed interval [*c*, *d*] is closed and bounded. Thus, *f*([*a*, *b*]) = [*c*, *d*]. ♦

## Connected Components

Let *X* be a topological space. For any two points *x*, *y* in *X*, write *x* ~ *y* if there’s a connected subset of *X* containing them. We claim this gives an equivalence relation; indeed, it is clearly reflexive and symmetry. Transitivity follows straight from proposition 3.

Definition. The equivalence classes of this relation are called theconnected componentsof X.

Note that each connected component *Y* is a *maximal* connected subset of *X*, in the sense that if we add any point outside *Y*, the resulting set won’t be connected.

We know that *X* is a set-theoretic disjoint union of its connected components. It’s tempting to think that *X* is a topological disjoint union as well. But that’s wrong. Indeed, let’s look at *X* = **Q** (as a subspace of **R**). The only connected subsets of *X* are the singleton points {*x*}: for if and with *a*<*b*, then we can pick an irrational *r*, *a*<*r*<*b* so that is a disjoint union of two non-empty open subsets. So *Y* is disconnected.

Thus, the connected components of **Q** are the singleton sets. But **Q** is clearly not discrete.

Definition. A space whose connected components are all singleton sets is said to betotally disconnected.

Thus, a discrete space is totally disconnected but not vice versa.

The example of **Q** shows that connected components are in general not open. However, they’re closed.

Proposition 5.Every connected component Y of a topological space X is closed.

**Proof**.

This follows almost immediately from proposition 2: since *Y* is connected, so is But *Y* is a maximal connected subset, so ♦

Proposition 6. If is a collection of non-empty topological spaces, the product is connected if and only if each is connected.

**Proof**.

(→) is obvious since each is the continuous image of the projection map from *X*. For (←), let be surjective and continuous. Let Since is open, it contains a basic open set of the form where for all but finitely many *i*‘s : Thus:

(#)

Next, use the following.

**Sublemma**. If*j*is an index, and satisfy: for all*i*except*i*=*j*, then they belong to the same connected component.**Proof**. The subspace is homeomorphic to*X*so it is a connected subset containing_{j}

Hence on the LHS of (#), we can change the coordinates one at a time and see that the whole is contained in i.e. *X* is connected. ♦

The above sublemma holds for the box topology too. This may con us into believing that under the box topology is also connected. But let’s take the space of all real sequences and *U* the subset of all sequences converging to 0. Any is also contained in which is open in the box topology and contained in *U*. Thus *U* is open. The same holds for any so *X*-*U* is also open. [ *Note that the sublemma implies altering a single term in a sequence has no effect on whether it converges to 0.* ]

## Examples

- An infinite set with the cofinite topology is connected since all non-empty closed subsets are finite and hence not open.
- The space is disconnected since and are open subsets of
*X*. These give the connected components of*X*. - By proposition 6, the square [0, 1] × [0, 1] is connected, as is its interior (0, 1) × (0, 1) in
**R**^{2}. - The circle
*S*^{1}is connected since it’s the continuous image of - (
**Topologist’s Sine Curve**) Take the set where Now*Y*and*Z*are homeomorphic to [0, 1] and (0, 1] respectively, so they’re connected. What’s surprising is that*X*is connected as well! Indeed, one observes that*Y*is a set of accumulation points for*Z*, since for every open point (0,*y*) in*Y*and ε>0, contains a point of*Z*. Thus Since*Z*is connected, proposition 2 tells us*X*is connected too.

**Exercise**.

Which of the following is/are correct for a subset *Y* of *X*?

- If cl(
*Y*) is connected, then so is*Y*. - If
*Y*is connected, then so is int(*Y*). - If int(
*Y*) is connected, then so is*Y*.

**Answer (Highlight to Read)**.

They’re all wrong! First statement: let *X* = **R** and *Y* be the union of (-1, 0) and (0, 1); cl(*Y*) = [-1, 1]. Second statement: let *X* = **R**^{2} and *Y* be the union of [-1, 0] × [-1, 0] and [0, 1] × [0, 1]. Each square is connected; since they share a common point, *Y* is connected. But int(*Y*) is the disjoint union of two open squares. Third statement: let *Y* be {0, 1} in **R**; int(*Y*) is empty and hence connected. ♦