Disjoint Union Topology - understanding operations on it

Question

I can find good explanations of how the disjoint union topology is constructed, but I am confused about how things such as complements, boundaries, limit points, etc. are to be understood in this context. For example, suppose we have two spaces, P and M and create their disjoint Union X with the disjoint union topology. It would seem that subsets of P and M must then be subsets of X that are disjoint. However, do they need to be separate as well or could a subset of P have limit points in a subset of M? With what open sets would the limit points be defined? How about the closure or boundary of unions of subsets of P and M? It seems from what I have been able to find that you could not define an open set in X that did not already exist in P or M, so I am confused. Any clarification or a pointer to a relevant treatment would be greatly appreciated.

Ernie

Answer 1

If $P$ and $M$ are disjoint topological spaces and $X = P\cup M$, then $X$ inherits a natural topology from $P$ and $M$, sometimes called the disjoint union topology. The open sets in this topology are all sets of the form $U\cup V$, where $U$ is open and $P$ and $V$ is open in $M$. In particular, since the empty set is open, any open subset of $P$ is open in $X$, and any open subset of $M$ is open in $X$.

The idea of this topology is that $P$ and $M$ form disconnected pieces of $X$, and do not interact in any way. Here are some basic properties:

• No sequence in $P$ or subset of $P$ has a limit point in $M$, and vice-versa.

• If $S\subset P$, then the closure of $S$ is also a subset of $P$. The same holds for $M$.

Answer 2

In this case both $P$ and $M$ are clopen (closed and open). So in particular the boundary of $P$ and $M$ is empty and no element of $P$ is a limit point of $M$ and vice-versa. On the other hand, an arbitrary subset $U$ of $X$ is open if and only if both $U\cap P$ and $U\cap M$ are open.

Answer 3

Another very useful property of $P \sqcup M$ is that for any space $X$ and any continuous functions $p:P \to X$, $m: M \to X$ , the unique function $f: P \sqcup M \to X$ which agrees with $p,m$ on $P,M$ respectively, is continuous. Thus the disjoint union is good for constructing continuous functions from it, which is a kind of dual to the product, which is good for constructing functions $ X \to P \times M$ in terms of its components.

Answer 4

A nice application of the disjoint union topology is it allows us to deduce the operator version of Kuratowski's closure-complement theorem (see Gardner and Jackson) as a corollary of the original theorem. In its original form, the theorem states that a maximum of 14 distinct subsets can be obtained by repeatedly applying the closure and complement operators to one subset in a topological space. The operator version states that a maximum of 14 distinct operators on a topological space can be obtained by repeatedly composing the closure and complement operators with the identity operator.

Since two operators $o_1$ and $o_2$ are distinct if and only if some subset $A$ satisfies $o_1A\neq o_2A,$ there exist topological spaces in which the number of distinct operators generated by closure and complement is strictly larger than the maximum number of distinct subsets any individual subset generates.* Hence, deducing the operator version from the original theorem requires something extra. That something turns out to be the disjoint union topology.

Suppose $X$ is a topological space in which $n$ distinct operators $o_1,\dots,o_n$ are generated by closure and complement. Thus, for each $1\leq i<j\leq n,$ there exists a subset $A_{ij}\subset X$ such that $o_iA_{ij}\neq o_jA_{ij}.$ Put the disjoint union topology on $$Y=\bigsqcup_{1\leq i<j\leq n}\!\!Y_{ij}$$ where $Y_{ij}=X$ for $1\leq i<j\leq n.$ For any subset $$B=\bigsqcup_{1\leq i<j\leq n}\!\!B_{ij}$$ of $Y$ and $1\leq m\leq n,$ it follows from the definition of the disjoint union topology that $$o_mB=\bigsqcup_{1\leq i<j\leq n}\!\!o_mB_{ij}.$$ Hence, the subset $$A=\bigsqcup_{1\leq i<j\leq n}\!\!A_{ij}$$ of $Y$ satisfies $o_iA\neq o_jA$ for all $1\leq i<j\leq n.$ It follows that $n\leq14$ by the closure-complement theorem (for subsets).

The operator version is usually proved directly (from which the subset version follows immediately), but the above construction illustrates that we can always find some subset in some space that distinguishes all the operators that get generated from an initial collection built from closure, complement, union, and intersection.

*When these two numbers are equal Gardner and Jackson call the space full.