Union of partitions
Let $X$ and $Y$ be sets such that $X\cap Y=\varnothing$. Let $\{A_\alpha\mid\alpha\in I\}$ and $\{B_\alpha\mid a\in J\}$ be partitions of $X$ and $Y$, respectively.
Prove that $\{A_\alpha\mid\alpha\in I\}\cup\{B_\alpha\mid a\in J\}$ is a partition of $X\cup Y$.
We have to check three conditions: a partition of a set $X$ is the set $\{A_\alpha\mid \alpha\in I\}$ of subsets $A_\alpha\subset X$ such that (1) every $A_\alpha$ is non-empty; (2) $\bigcup_{\alpha\in I}A_\alpha=X$; (3) for every $\alpha,\beta\in I$ we have $A_\alpha \cap A_\beta=\varnothing$ or $A_\alpha=A_\beta$.
For the first condition, $\{A_\alpha\mid\alpha\in I\}$ and $\{B_\alpha\mid a\in J\}$ are non-empty by definition, so the union is also non-empty.
However, I don't know how to prove the second and third condition.
I have the same problem with proving the second condition in the following problem: $L_y=\{(x,y)\in\mathbf{R}^2 \mid x\in\mathbf{R}\}$ with $y\in\mathbf{R}$ is a partition of $\mathbf{R}^2$..
Could someone help?
$\endgroup$ 22 Answers
$\begingroup$Let $\{A_\alpha\mid \alpha\in I\}$ be a partition of $X$ and let $\{B_\alpha\mid \alpha \in J\}$ be a partition of $Y$ where $X\cap Y=\emptyset$.
We wish to prove that $\bigcup \left(\{A_\alpha\mid\alpha\in I\}\cup\{B_\alpha\mid\alpha\in J\}\right)=X\cup Y$
Method 1:
We may assume without loss of generality that $I\cap J=\emptyset$. (If not, then simply replace $J$ with something that is actually disjoint from $I$ and rename all of the labels of the parts in $\{B_\alpha\mid \alpha\in J\}$ to match the changes made).
While we're at it, let us also rename all of the sets in $\{B_\alpha\mid\alpha\in J\}$ to be instead $\{A_\alpha\mid\alpha\in J\}$. Since $I\cap J=\emptyset$ now, there should be absolutely no ambiguity between the sets in the partition of $X$ and the sets in the partition of $Y$.
We have then $$\bigcup\limits_{\alpha\in (I\cup J)}A_\alpha=\left(\bigcup\limits_{\alpha\in I}A_\alpha\right)\cup \left(\bigcup\limits_{\alpha\in J}A_\alpha\right)=X\cup Y$$
Method 2:
If the previous method with all of my label changing was unsatisfactory, we can prove this a different way with element chasing.
The union of all sets included in one of $\{A_\alpha\mid \alpha\in I\}$ or $\{B_\alpha\mid\alpha\in J\}$ will of course be a subset of $X\cup Y$ as each $A_\alpha$ is a subset of $X$ and each $B_\alpha$ is a subset of $Y$.
The reverse inclusion can be shown in the following way: Take an arbitrary element of $X\cup Y$. Let's call it $z$. One of two things will happen. Either $z\in X$ or $z\in Y$.
In the first case where $z\in X$ we have since $\{A_\alpha\mid \alpha\in I\}$ is a partition of $X$, there must be some $\alpha$ for which $z\in A_\alpha$. It follows then that since that very same $A_\alpha$ is included as one of the sets in the collection $\{A_\alpha\mid \alpha\in I\}\cup \{B_\alpha\mid \alpha\in J\}$ that $z$ is also in the union of the sets in that collection.
If this still isn't clear, then look at an example. Let $X=\{1,2,3,4,5\}$ and $Y=\{a,b,c,d,e\}$. Let the partition of $X$ be $\{\{1,2,3\},\{4,5\}\}$ and let the partition of $Y$ be $\{\{a,b\},\{c,d\},\{e\}\}$
We have then $\{\overbrace{\underbrace{\{1,2,3\},\{4,5\}}_{\text{covers}~X},\underbrace{\{a,b\},\{c,d\},\{e\}}_{\text{covers}~Y}}^{\text{covers}~X\cup Y}\}$. My first method tried to generalize this idea that we could still organize the one collection together that formed the partition of $X$ separately from the collection that formed the partition of $Y$. My second method looked a bit more closely at the individual elements and checked that they were in fact all still there somewhere.
The third condition, that two distinct sets in the collection must be disjoint, either the two distinct sets will be parts from the same partition and therefore disjoint, or one will be a subset of $X$ and the other of $Y$ and therefore disjoint since $X$ and $Y$ are disjoint.
The second problem is very similar. Just note that every element of $\Bbb R^2$ can be described first by the $y$ coordinate. This tells you which part it will belong to. That the parts are disjoint follows from the fact that the $y$ coordinates of elements from two distinct parts will necessarily be different.
$\endgroup$ 1 $\begingroup$This is a place where indices such as $\alpha\in I$ just make things more difficult.
A partition $\mathscr{C}$ on the set $Z$ is a subset of $P(Z)$ such that
- $\emptyset\notin\mathscr{C}$
- for all $C_1,C_2\in\mathscr{C}$, if $C_1\ne C_2$ then $C_1\cap C_2=\emptyset$
- $\bigcup\mathscr{C}=Z$
Now you have a partition $\mathscr{A}$ of $X$ and a partition $\mathscr{B}$ of $Y$, with $X\cap Y=\emptyset$, and you wish to show that $\mathscr{C}=\mathscr{A}\cup\mathscr{B}$ is a partition of $Z=X\cup Y$.
Since $\emptyset\notin\mathscr{A}$ and $\emptyset\notin\mathscr{B}$, w have $\emptyset\notin\mathscr{C}$.
Suppose $C_1,C_2\in\mathscr{C}$ and $C_1\ne C_2$. There are four cases
- $C_1\in\mathscr{A}$ and $C_2\in\mathscr{A}$
- $C_1\in\mathscr{A}$ and $C_2\in\mathscr{B}$
- $C_1\in\mathscr{B}$ and $C_2\in\mathscr{A}$
- $C_1\in\mathscr{B}$ and $C_2\in\mathscr{B}$
In cases 1 and 4, we conclude that $C_1\cap C_2=\emptyset$ because they are distinct members of a partition; in cases 2 and 3 we conclude that $C_1\cap C_2=\emptyset$ because they're subsets of disjoint sets
Since $\bigcup\mathscr{C}=\Bigl(\bigcup\mathscr{A}\Bigr)\cup\Bigl(\bigcup\mathscr{B}\Bigr)=X\cup Y$ we have $\bigcup\mathscr{C}=Z$.
