## Disjoint Union Topology

Before we move onto the next topic in our discussion of manifolds I think it would be nice to pause and discuss a previous left out topic. Just as was the case for the product and subspace topologies the *disjoint union topology *(defined below). But, before we actually define the disjoint union topology let us first describe the disjoint union.

**Disjoint Union:** Let be a non-empty collection of non-empty sets and define the *disjoint union* of the collection as where .

The disjoint union can be thought of as taking a collection of sets and uniting them but keeping the lineage (where the elements came from) apparent. It can be thought of as painting each element of each a different color. It should be noted though that there is no redundancy of elements in this definition. In the sense that then

So, now that we know what the disjoint union is let us define the disjoint union topology (D.U.T.)

**Disjoint Union Topology (D.U.T.):** Let be a non-empty collection of non-empty topological spaces. Define the *canonical injection * where by . Then, topologize by calling open if and only if is open for every . With this topology we call the *disjoint union topology (D.U.T.) *

A universal characterization of the D.U.T. (I won’t go full into this…despite the name of the forum this isn’t category theory…yet) is the following.

**Theorem:** A function is continuous if and only if is continuous for each .

**Proof:** Suppose that is continuous and let be open. Then, is open in and so is open in for each . Thus, is continuous.

Conversely, suppose that is continuous for each and let be open. Then, is open in since is open for each .

The conclusion follows. .

The next theorem gives an idea of what the D.U.T. looks like on a class of homeomorphic spaces.

**Theorem:** Let be a class of topological spaces such that with the homeomorphism. Then where has the D.U.T. and the discrete.

**Proof:** Define by .

We first prove that is a bijection. So, suppose that then . It is immediately obvious that . Thus, since is injective . To see that it’s surjective we must merely note that if then and so for some and so .

We now prove that is continuous. To see this let be open. Then where is open in and .

We claim that

To see this first let then and so . Thus, thus and so .

Conversely, let then and thus and so by definition

It easily follows that is a union of open sets in .

A similar (and equally tedious) calculation shows that is open, we omit it here. The conclusion follows.

It should be obvious that the canonical injections are both open and closed and thus embeddings.

**Theorem:** Let be a class of discrete spaces. Then, under the D.U.T. is discrete.

**Proof:** Let then

It follows that is open in for each . Thus, is open. The conclusion follows. .

Next we prove some separation invariants.

**Theorem: **Let be a non-empty collection of Kolomogorov spaces. Then, under the D.U.T. is Kolomogorov.

**Proof:** Let be distinct. If then clearly and are disjoint neighborhoods of each (must stronger than what is needed). If then must be distinct. But, since is Kolomogorov there exists, WLOG, a neighborhood which contains but not . Then, clearly which is open and . The conclusion follows. .

**Theorem:** Let be a non-empty class of spaces. Then, under the D.U.T. is .

**Proof:** Let be distinct. Just as last time if then and are the desired neighborhoods. So, assume not. Then, are distinct and thus by assumption there exists neighborhoods and which contain and do not contain respectively. Clearly then and are the desired neighborhoods. .

And lastly, most interesting for our current topic of study.

**Theorem:** Let be a non-empty collection of Hausdorff spaces then under the D.U.T. is Hausdorff.

**Proof:** Let be distinct. If then and are exactly the neighborhoods of that we need. So assume that , then are distinct. It follows from ‘s assumed Hausdorffness that there exists disjoint neighborhoods containing respectively. Clearly then and are the desired neighborhoods. .

We now discuss countability invariants of the D.U.T.

**Theorem:** Let be a non-empty collection of first countable spaces then is first countable under the D.U.T.

**Proof:** Let and let be the guaranteed countable base at . Define . Clearly these are open in . To see that it’s a base at let be a neighborhood of in . Thus, is a neighborhood of in . Thus, there exists some such that . It follows that and since the conclusion follows.

Once again our next theorem will be important for the main theorem of this section.

**Theorem: **Let be a non-empty collection of second countable spaces. Then, is second countable if is countable.

**Proof:** Let be the class of countable bases in each and let where . Clearly then being the countable union of countable sets is countable. Also, each element of is clearly open. Thus, it remains to show that is a base. To do this let be arbitrary and any neighborhood of it. Clearly then is a neighborhood of in . Thus, we may find some such that . It follows that . Since the conclusion follows.

We are now at the final part of our list theorems and it should be apparent now what the piece de resistance shall be.

**Theorem:** Let be a non-empty class of locally Euclidean spaces of dimension . Then, is locally Euclidean of dimension .

**Proof:** Let be arbitrary. By assumption of being locally Euclidean there exists some chart at . Clearly then is a neighborhood of . Define a mapping by .

Clearly this is a bijection since

It is surjective since implies that which implies that which then of course implies that .

Now, it is clearly continuous since is open implies that is open in and thus since is open it is open in . Thus, is open in and thus lastly is open in . Similarly, is open implies that is open in and thus is open in and thus finally is open in .

Remembering that is open in we may conclude that is a chart at . The conclusion follows by noting the arbitrariness of . .

We are now prepared to state our main theorem.

**Theorem: **Let be countable collection of -manifolds. Then, is an -manifold.

**Proof: **Combining our last couple theorem the above is justs a result since the countable union of a second countable Hausdorff locally Euclidean space of dimension is second countable Hausdorff and locally Euclidean of dimension . .

Now that we have proved the main result for this post we prove some ancillary theorems regarding the D.U.T.

**Theorem:** Let be finitely many topological spaces, then is compact if and only if is for each .

**Proof:** First suppose that is compact and let be an open cover of clearly then is an open cover for and so by assumption it has a finite subcover . It clearly follows that if then covers and so is a finite subcover of .

Conversely, remember that the canonical injection is an embedding and thus we must only show that is compact. To see this note that is the union of all the . It is easy to see that this is open since

and thus is closed. Thus, $ is a closed subspace of a compact space and thus compact. It follows from previous comment that is compact.

The conclusion follows.

It should come as a surprise to no one that disjoint unions react very badly with connectedness.

**Theorem:** Let be a collection of (at least two) non-empty topological spaces. Then, is disconnected.

**Proof:** Clearly as was stated twenty times before and it’s open since

But it is also closed since

The conclusion readily follows.

This concludes our “short” introduction to the disjoint union topology.

[…] spaces. Then, the coproduct of and is the set (the formal disjoint union of and ) with the disjoint union topology and the canonical injections, is the coproduct of and in […]

Pingback by Categorical Coproducts « Abstract Nonsense | February 7, 2012 |

[…] second way we can construct new manifolds out of old ones is via the disjoint union of topological spaces. Indeed, suppose that we have two -dimensional topological manifolds and with atlases and […]

Pingback by Topological Manifolds (Pt. II) « Abstract Nonsense | August 30, 2012 |