## Disjoint Unions

Let *X* and *Y* be topological spaces and be a set-theoretic disjoint union. We wish to define a topology on *Z* in a most natural way.

Definition. The topology on is defined to be:

It’s almost trivial to check that this satisfies the axioms of a topology. E.g. to check *T* is closed under union of arbitrarily many elements, we have

Pictorially, the definition for *T* is quite straight-forward. One imagines placing the two spaces side-by-side and separate them by an insurmountable gulf.

More generally, one can take the disjoint union of an arbitrary collection of topological spaces , where the open subsets of *U* are precisely those of the form for open subsets .

The bases and subbases of *X _{i}* give rise to that of

*X*in a natural manner:

Proposition. Let be a collection of topological spaces and .

- If is a basis of for each i, then is a basis for X.
- If is a basis of for each i, then is a basis for X.

**Proof (of first property)**

Let *T* be the topology for *X* obtained via disjoint union. Clearly, each since an open subset of *X _{i}* is also open in

*X*. Thus Conversely, every open subset

*U*of

*X*is a (disjoint) union of open subsets of

*X*, each of which is a union of elements of . Hence

_{i}*U*is a union of elements from

*B*and

*B*is a basis for

*T*. ♦

The proof for the second property is similar and we won’t repeat ourselves. Since the disjoint union can be thought of as “addition” between topological spaces, it shouldn’t surprise us to find the distributive property for product.

Theorem (Distributive Property). Let be a collection of topological spaces. Then:

**Proof**.

We clearly have a set-theoretical bijection between both sides. It remains to see that the collection of open subsets on both sides correspond.

Now, a basis of the LHS is given by for some open subsets , . But this is just and since each is open, so is their (disjoint) union. Thus, the LHS topology is contained in the RHS.

Conversely, a basis of the each term of the RHS is given by , where , are open. By the above proposition, the set is a basis of RHS. Since each element of this basis is open in LHS, the RHS topology is contained in the LHS. ♦

**Note**.

The fact that multiplication is distributive over an *infinite* sum should not be taken for granted. Indeed, the corresponding case for algebra requires careful consideration of convergence on both sides.

Corollary. Let X be a topological space. If I is an index set, endowed with the discrete topology, then X × I is isomorphic to a disjoint union of , where each .

**Proof**.

Consider the singleton set {*i*} with the obvious topology. Now *I* is the disjoint union of {*i*}, over . Hence:

where each component in the rightmost expression is isomorphic to *X*. ♦

## Metrisability of Disjoint Unions

Next, we’re interested in the following question.

If X and Y are metrisable topological spaces, is metrisable too?

On an intuitive level, the space *Z* comprises of components *X* and *Y* such that points in different components are extremely far from each other. But one can’t set the distance between them to be infinite, since the metric can only take real values. [ One might understandably think of extending the metric function to include infinite distances, or even “nonstandard reals”, but that opens up a whole new can of worms. ]

However, it turns out that if *X* is a metrisable topological space, then we can pick a metric such that *d*(*x*, *y*) < 1 for any *x*, *y* in *X*.

Proposition. If (X, d) is a metric space, then the function , , is also a metric.

**Proof**.

It suffices to show that if *r*, *s*, *t* are non-negative real numbers satisfying , then . Since is a strictly increasing function, we may assume *t* = *r*+*s*. Then clearing denominators gives:

♦

Hence, if each *X _{i}* is a metrisable topological space, then we can pick a metric

*d*which is bounded by 1. [ We say that a metric

_{i}*d*on

*X*is

**bounded**by

*B*if

*d*(

*x*,

*y*) ≤

*B*for any

*x*,

*y*in

*X*. ] Now we just define a metric on the disjoint union via:

for , we have

Let’s summarise what we’ve learnt in the last few paragraphs.

- If
*X*is a metrisable metric space, then we may pick a metric which is bounded by 1. In other words,*the condition that a metric is bounded is not a topological invariant*. - If each
*X*is a metrisable topological space, then so is their disjoint union._{i}

## Connected Spaces

On the other hand, let’s start with some generic topological space *X*. We’d like to partition *X* as a disjoint union of topological subspaces. Note that if , then each is an open subset. The following theorem shows that this suffices to characterise *X* as a disjoint union.

Theorem. Suppose , where each is open and for any i≠j. Then .

**Example**

Consider the subset of **R**. Then where , and . Each of these sets is open in *Y* since

, and

Since they form a disjoint union of *Y*, we have .

**Proof**.

From the inclusion maps , we get a set-theoretic map . The map is surjective since *X* is the union of *U _{i}*‘s. It is injective since any two distinct

*U*‘s don’t intersect.

_{i}To show that *f* is a homeomorphism, each open subset of is of the form for open . Since is also open, we see that is open. Thus is open in *X*.

Conversely, if *V* is an open subset of *X*, then , where is open in *U _{i}*. Thus is open in ♦

Note that in the theorem, each *U _{i}* is also closed since it’s complement is which is open. Thus the

*U*are clopen subsets of

_{i}*X*: if

*X*is a disjoint union of more than one components, than each component is a clopen subset.

Conversely, suppose is clopen such that . Then satisfies the conditions of the above theorem, so . Hence, we have the following.

Corollary. If X has no non-trivial clopen subsets, then we cannot write in a non-trivial manner. When that happens, we say that X isconnected.

In a nutshell, connected spaces are those which cannot be broken up any further into disjoint unions. We’ll be looking at these in greater detail later.