Putting a topology on the disjoint union and what are the open sets.

Question

I have the following question:

Could anyone give me hints to answer those questions ?

Answer 1

Given two spaces X and Y, the disjoint union of X and Y,
X + Y = X×{1} $\cup$ Y×{2}.
The elements of x in X and y in Y are respectively tagged
(x,1) and (y,2) to insure X and Y are indeed disjoint.

For example, the disjoint union of [0,1] and [0,1] is [0,1]×{1,2}.
Basically two parallel closed intervals.

The topology of X + Y is
{ U×{1} $\cup$ V×{2} : U open within X, V open within Y }.
The disjoint union can likewise be defined for any number of spaces.
Even infinite disjoint unions.

Exercise. Prove the following.
The topology given for X + Y is indeed, a topology.
X×{1} and Y×{2} are clopen.
{ U×{1}, V×{2} : U open within X, V open within Y }
is a base for X + Y.
A refinement of this result using a base for X and a base for Y
instead of the topologies of X and Y.
X + Y and Y + X are homeomorphic.

A generalization of these results for any number of spaces.
X + (Y + Z) is homeomorphic to X + Y + Z.

Intuitively, disjoint unions are different planets, parallel universes.

Answer 2

Questions a) and b) are the same. A topology on $X$ is a set $T$ of subsets of $X$, which are called the open sets of $X$.

To be a topology, $T$ must satisfy these axioms:

1. $\emptyset \in T$ and $X \in T$.
2. $T$ is stable under finite intersection : if $U_1,\dots,U_n \in T$, then $\bigcap_{i=1}^n U_i \in T$.
3. $T$ is stable under union : if $(U_i)_{i\in I} \subseteq T$, then $\bigcup_{i\in I} U_i \in T$.

Now, you have a topology $T$ on $X$ and a topology $T'$ on $Y$. Let's start with a naive guess for the topology $T^*$ on $X \sqcup Y$ : we chose $T \cup T'$.

$T^*$ is not a topology on $X\sqcup Y$ since $X\sqcup Y \notin T^*$

Let's say we add this missing piece, $T^* := T \cup T' \cup \{X \sqcup Y\}$, then $T^*$ still isn't a topology since axiom 3. above is not fulfilled in general (take $U \in T$ and $V \in T'$, then $U \cup V$ might not be in $T^*$).

Let's make it work by force :

$$\boxed{T'' := \{U \sqcup V \ \big| \ U \in T, \, V \in T'\}}$$

With this definition, $T''$ satisfies axioms 1. and 3. above (prove it!). I leave it to you to show that it also satisfies axiom 2.