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One way of distinguishing between different topological spaces is to look at the way thay "split up into pieces".

To make this idea rigorous we need the idea of *connectedness*.

**Definition**

A space which is a union of two disjoint non-empty open sets is called **disconnected**.

**Equivalently**

A space *X* is **connected** if the only subsets of *X* which are both open and closed (= *clopen*) are and *X*.

**Proof of equivalence**

If *X* = *A* *B* with *A* and *B* open and disjoint, then *X* - *A* = *B* and so *B* is the complement of an open set and hence is closed. Similarly, *B* is clopen.

Conversely, if *A* is a non-empty, proper open subset then *A* and *X* - *A* disconnect *X*.

A *subset* of a topological space is called connected if it is connected in the subspace topology.

**Example**

The set [0, 1] [2, 3] **R** with its usual topology is not connected since the sets [0, 1] and [2, 3] are both open in the subspace topology.

**Theorem**

*The interval* (0, 1) **R** *with its usual topology is connected.*

**Proof**

Suppose that (0, 1) = *A* *B* with *A*, *B* disjoint non-empty clopen subsets. Choose *a* *A* and *b* *B* with (say) *a* < *b*.

Then let be the least upper bound of the set *C* = { ([*a*, *b*] *A* }. This least upper bound exists by the standard properties of **R**.

Since *C* is a closed subset it contains its limit points and so *C* and hence is in *A*. Since *A* is open has an -neighbouhood lying inside *A* and so unless = *b* it would not be an upper bound of *C*. But = *b* contradicts the fact that *b* *B* = (0, 1) - *A*.

A similar proof shows that any interval is a connected subset of **R**. In fact, we have:

**Theorem**

*Intervals are the only connected subsets of* **R** *with the usual topology.*

**Proof**

If *A* **R** is not an interval, then choose *x* **R** - *A* which is not a bound of *A*.

Then the subsets *A* (-, *x*) and *A* (*x*, ) are open subsets in the subspace topology *A* which would disconnect *A* and we would have a contradiction.

The most important property of connectedness is how it affected by continuous functions.

**Theorem**

*The continuous image of a connected space is connected.*

**Proof**

If *f*: *X* *Y* is continuous and *f*(*X*) *Y* is disconnected by open sets *U*, *V* in the subspace topology on *f*(*X*) then the open sets *f* ^{-1}(*U*) and *f* ^{-1}(*V*) would disconnect *X*.

**Corollary**

*Connectedness is preserved by homeomorphism.*

We may use this fact to distinguish between some non-homeomorphic spaces.

**Example**

The spaces [0, 1] and (0, 1) (both with the subspace topology as subsets of **R**) are not homeomorphic.

**Proof**

Removing any point from (0, 1) gives a non-connected space, whereas removing an end-point from [0, 1] still leaves an interval which is connected.

A similar method may be used to distinguish between the non-homeomorphic spaces obtained by thinking of the letters of the alphabet as in Exercises 1 question 1.

**Corollary** (**The Intermediate Value Theorem**)

*If f*: **R** **R** *is continuous then for any a, b in* **R***, f attains any value between* *f*(*a*) *and* *f*(*b*) *at some point between a and b*.

**Proof**

Since the interval [*a*, *b*] is connected, so is its image *f*([*a*, *b*]) and so this too is an interval.

**Remark**

A similar result holds for a continuous real-valued function on *any* connected space.

Fastening together connected space "with an overlap" gives a conected space:

**Theorem**

*If A and B are connected and A B then A B is connected*.

**Proof**

Suppose open sets *U* and *V* disconnect *A* *B*. Then *A* *U* and *A* *B* would disconnect *A* and so one of them is . So suppose *A* *U*.

Similarly we have either *B* *V* or *B* *U*. Since *B* meets *A* the first of these is imposssible and so we have *A* *B* *U* and *V* = .

In fact if {*A*_{i}| *i* *I*} is any set of connected subsets with *A*_{i} then *A*_{i} is connected.

**Definition**

The maximal connected subsets of a space are called its **components**.

Note that every point of a space lies in a *unique* component and that this is the union of all the connected sets containing the point (This is connected by the last theorem.)

**Examples**

- The components of the space [0, 1] [2, 3] with the subspace topology inherited from
**R**, are the subspaces [0, 1] and [2, 3].

- Components of
**Z**(with the subspace topology from**R**) are the single points.

- (Harder!) Components of
**Q**(with the subspace topology from**R**) are also single points.**Proof**

If*r*<*s*are rationals, choose an irrational*x*between them. Then**Q**(-,*x*) and**Q**(*x*, ) disconnect**Q**and so*r*,*s*are in different components.

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