Topology: Disjoint Unions

Disjoint Unions

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

Definition. The topology on $Z := X \coprod Y$ is defined to be:

$T = \{U \coprod V : U \subseteq X, V \subseteq Y \text{ open }\}.$

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

$\cup_i (U_i \coprod V_i) = (\cup_i U_i)\coprod (\cup_i V_i) \in T.$

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 $X = \coprod_i X_i$ , where the open subsets of U are precisely those of the form $U = \coprod_i U_i$ for open subsets $U_i\subseteq X_i$ .

The bases and subbases of X_i give rise to that of X in a natural manner:

Proposition. Let $X_i$ be a collection of topological spaces and $X := \coprod_i X_i$ .

If $B_{X_i}$ is a basis of $X_i$ for each i, then $B:=\coprod_i B_{X_i}$ is a basis for X.

If $S_{X_i}$ is a basis of $X_i$ for each i, then $S:=\coprod_i S_{X_i}$ is a basis for X.

Proof (of first property)

Let T be the topology for X obtained via disjoint union. Clearly, each $B_{X_i}\subseteq T$ since an open subset of X_i is also open in X. Thus $B\subseteq T.$ Conversely, every open subset U of X is a (disjoint) union of open subsets of X_i, each of which is a union of elements of $B_{X_i}$ . Hence 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 $X, Y_i$ be a collection of topological spaces. Then:

$X \times (\coprod_i Y_i) \cong \coprod_i (X\times Y_i).$

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 $U\times (\coprod_i V_i)$ for some open subsets $U\subseteq X$ , $V_i\subseteq Y_i$ . But this is just $\coprod_i (U\times V_i)$ and since each $U\times V_i\subseteq X\times Y_i$ is open, so is their (disjoint) union. Thus, the LHS topology is contained in the RHS.

Conversely, a basis of the each term $X\times Y_i$ of the RHS is given by $\{U\times V_i\}$ , where $U\subseteq X$ , $V_i\subseteq Y_i$ are open. By the above proposition, the set $\cup_i \{U\times V_i : U\subseteq X, V_i\subseteq Y_i\}$ 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 $\{X_i\}_{i\in I}$ , where each $X_i \cong X$ .

Proof.

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

$X \times I \cong X\times \left(\coprod_{i\in I} \{i\}\right)\cong \coprod_{i\in I} \left(X \times \{i\}\right),$

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 $Z = X\coprod Y$ metrisable too?

On an intuitive level, the space Z comprises of components X and Y such that points $x\in X, y\in Y$ 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 $d' : X\times X\to \mathbf{R}$ , $d'(x, y) = \frac{d(x,y)}{1+d(x,y)} < 1$ , is also a metric.

Proof.

It suffices to show that if r, s, t are non-negative real numbers satisfying $r+s\ge t$ , then $\frac{r}{1+r}+\frac{s}{1+s} \ge \frac{t}{1+t}$ . Since $x\mapsto \frac{x}{1+x}$ is a strictly increasing function, we may assume t = r+s. Then clearing denominators gives:

$r(1+s)(1+r+s)+s(1+r)(1+r+s) - t(1+r)(1+s) = r^2 s + s^2 r + 2rs \ge 0.$ ♦

Hence, if each X_i is a metrisable topological space, then we can pick a metric d_i which is bounded by 1. [ We say that a metric 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 $X := \coprod_i X_i$ via:

for $x_i\in X_i, x_j \in X_j$ , we have $d(x_i, x_j) = \begin{cases}d_i(x_i, x_j), &\text{ if } i=j,\\ 2,&\text{ if } i\ne j.\end{cases}$

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_i is a metrisable topological space, then so is their disjoint union.

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 $X:= \coprod_i X_i$ , then each $X_i\subseteq X$ is an open subset. The following theorem shows that this suffices to characterise X as a disjoint union.

Theorem. Suppose $X = \cup_i U_i$ , where each $U_i\subseteq X$ is open and $U_i\cap U_j=\emptyset$ for any i≠j. Then $X \cong \coprod_i U_i$ .

Example

Consider the subset $Y=(-\infty, -1]\cup\{0\}\cup [1,\infty)$ of R. Then $Y = U \cup V \cup W$ where $U=(-\infty, -1]$ , $V = \{0\}$ and $W = [1,\infty)$ . Each of these sets is open in Y since

$U=(-\infty, -1/2)\cap Y$ , $V=(-1/2,+1/2)\cap Y$ and $W=(1/2,\infty)\cap Y.$

Since they form a disjoint union of Y, we have $Y \cong U\coprod V\coprod W$ .

Proof.

From the inclusion maps $U_i\hookrightarrow X$ , we get a set-theoretic map $f:\coprod_i U_i \to X$ . The map is surjective since X is the union of U_i‘s. It is injective since any two distinct U_i‘s don’t intersect.

To show that f is a homeomorphism, each open subset of $\coprod_i U_i$ is of the form $\coprod_i V_i$ for open $V_i\subseteq U_i$ . Since $U_i\subseteq X$ is also open, we see that $V_i\subseteq X$ is open. Thus $f(\coprod V_i)=\cup_i V_i$ is open in X.

Conversely, if V is an open subset of X, then $f^{-1}(V) = \coprod_i V_i$ , where $V_i = V\cap U_i$ is open in U_i. Thus $f^{-1}(V) = \coprod_i V_i$ is open in $\coprod_i U_i.$ ♦

Note that in the theorem, each U_i is also closed since it’s complement is $\cup_{j\ne i} U_j$ which is open. Thus the U_i are clopen subsets of X: if X is a disjoint union of more than one components, than each component is a clopen subset.

Conversely, suppose $U\subseteq X$ is clopen such that $U\ne X,\emptyset$ . Then $X = U \cup (X-U)$ satisfies the conditions of the above theorem, so $X \cong U \coprod (X-U)$ . Hence, we have the following.

Corollary. If X has no non-trivial clopen subsets, then we cannot write $X \cong Y_1 \coprod Y_2$ in a non-trivial manner. When that happens, we say that X is connected.

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.