Prove $\{A_1, \cdots ,A_n\}$ is a partition of $A$ given $\{S_1, \cdots, S_n\}$ is a partition of $\Omega$ Let $A \subseteq \Omega$ and $\{S_1,\cdots ,S_n\}$ be a partition of $\Omega$. Let $A_i = A \cap S_i$. Prove $\{A_1, \cdots ,A_n\}$ is a partition of $A$.
I'm having trouble formalizing this with mathematical symbols but I think I get the idea behind the problem...
To prove something is a partition we must show that is satisfies the conditions:


*

*1) The subsets are mutually disjoint

*2) The union of the subsets is equal to the whole sample space


We are given that $\{S_1,\cdots ,S_n\}$ is a partition of the sample space $\Omega$ so we know the collection of $S_i$ are mutually disjoint and their union $=\Omega$
We can also view this intuitively by the following example. Let $\Omega$ be a piece of lined paper. Let the lines on this paper represent the partitions $S_i$ with $1\le i \le n$ where $n$ equals the number of lines. Then draw a circle on this piece of paper and no matter what circle you draw (no matter how imperfect the shape) the lines will always divide the circle into mutually disjoint partitions $A_i$ that all together add up to give the full circle.
There are two cases for each section separated by lines:


*

*Case 1 - No part of the circle is inside the section meaning $A_i\notin S_i$

*Case 2 - Some or all of the circle is inside the section meaning $A_i \in S_i$


How do I put all of this logic together in symbols to write it in a concise, mathematical way?
Thank you!
 A: Clearly

$$A_i\cap A_j= \{x\in \Omega | x\in A \land x\in S_i \land x\in S_j\}=A\cap S_i\cap S_j=\begin{cases} A_i & i=j \\ \varnothing & i\ne j\end{cases}.$$

So the sets are disjoint.
Now let $x\in A$. Then as $A\subseteq \Omega$, $\exists 1\le i\le n$ so that $x\in S_i$ because $\{S_i\}_{i=1}^n$ is a partition of $\Omega$, in logical symbols

$$\Omega = \bigcup_{i=1}^nS_i=\{x\in\Omega | \exists 1\le i\le n \text{ such that } x\in S_i\}.$$

Because $x\in A \land x\in S_i$, we have $x\in \{x\in \Omega | x\in A\land x\in S_i\}=A\cap S_i=A_i$. Hence $\{A_i\}_{i=1}^n$ is a partition of $A$ by definition.
A: 
I'm having trouble formalizing this with mathematical symbols but I think I get the idea behind the problem...

Frame challenge: we don't need symbols for a formal mathematical proof.
First, we want to prove that the $A_i$ are disjoint. Consider two of them, $A_i$ and $A_j$. Being disjoint means "if a point is in one of them, then it's not in the other". So we're going to prove that. Let $x$ be a point in $A_i$.
We know that $A_i$ is the intersection of $A$ and $S_i$, which in particular means that $A_i$ is a subset of $S_i$. So, if $x$ is a point of $A_i$, then $x$ is also a point of $S_i$. But we know that $S_i$ and $S_j$ are disjoint, so that means that $x$ is not a point of $S_j$. And we also know that $A_j$ is a subset of $S_j$, so if $x$ is not a point of $S_j$, then $x$ is not a point of $A_j$.
Thus we have proven that for any point $x$ in $A_i$, $x$ cannot be a point in $A_j$. This proves that $A_i$ and $A_j$ are disjoint. This is true for any pair $A_i$ and $A_j$, so the whole family $A_1, \ldots, A_n$ is pairwise-disjoint.
Second, we want to prove that the union of the $A_i$ is all of $A$. In other words, we want to prove that for any point $x$ in $A$, there exists an $i$ such that $x$ is a point of $A_i$. So we're going to prove that. Let $x$ be a point of $A$.
$A$ is a subset of $\Omega$, so if $x$ is a point of $A$, then $x$ is also a point of $\Omega$. We know that the union of the $S_i$ is all of $\Omega$, so there exists an $i$ such that $x$ is a point of $S_i$. So, $x$ is a both in $S_i$ and in $A$: this is the definition of being in their intersection $S_i \cap A$. But we know that $A_i = S_i \cap A$. So we have proven that $x$ is in $A_i$.
Thus we have proven that for any point $x$ in $A$, there exists an $i$ such that $x$ is in $A_i$. This proves that the union of the $A_i$ covers $A$.
We have proven the two conditions: the family of subsets $(A_1, \ldots, A_n)$ is indeed a partition of $A$.

We can also view this intuitively by the following example. Let $\Omega$ be a piece of lined paper.

Pardon the flashy colours. Here is a visual of partition $(S_1, S_2, \ldots, S_n)$ of $\Omega$, and the induced partition $(A_1, A_2, \ldots, A_n)$ of $A$:

