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I have the following question:

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Could anyone give me hints to answer those questions ?

Math1000
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Emptymind
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    Why not just take the union of their corresponding topologies as a basis? – Rushabh Mehta Jan 22 '20 at 08:00
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    Declare a set to be open if its intersection with $X$ is open in $X$ and ts intersection with $Y$ is open in $Y$ . – Kavi Rama Murthy Jan 22 '20 at 08:02
  • what do you mean by "their corresponding topologies"? @DonThousand – Emptymind Jan 22 '20 at 08:02
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    A topological space is a set with a topology. Take the union of the topologies, and that's your basis. – Rushabh Mehta Jan 22 '20 at 08:03
  • @KaviRamaMurthy subspace topology? why? – Emptymind Jan 22 '20 at 08:03
  • @Emptymind No. What Kavi's suggesting is the other way around. However, it is true that this topology is compatible i.e., the subset topology of $X$ in $X\coprod Y$ becomes the topology we started with. – WoolierThanThou Jan 22 '20 at 08:04
  • @WoolierThanThou could you please explain your comment in more details? – Intuition Jan 26 '20 at 00:35
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    Think of it this way: How many topologies are there on $X\coprod Y$ such that the subspace of topology on $X$ respectively $Y$ become the topology they started with? Well, since any $A=A\cap X \coprod A\cap Y$, we see that the disjoint union topology is the only such topology on $X\coprod Y$. – WoolierThanThou Jan 26 '20 at 00:48

2 Answers2

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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.

William Elliot
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  • Could you please look at this question (if you have time) https://math.stackexchange.com/questions/3518158/the-sum-of-two-sets-and-the-disjoint-union ? – Emptymind Jan 23 '20 at 08:41
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    It seems @Emptymind to be stating that the topology of S + T is the final topology of the embedding of S and T into S + T. That diagram thing and its fundamenties is just obscuring the problem. Thus I ignored it. What is the final topology of the embeddings of S and T into S + T wouldn't have been discarded as neojargon. However, the current problem is just the result of the final topology. – William Elliot Jan 23 '20 at 13:00
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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.

Olivier Roche
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