Let $\sim_1$ and $\sim_2$ be distinct equivalence relations on $A$. Define $\sim_3$ by $a\sim_1 b$ and $a\sim_2 b$. Let $[x]_i$ denote the equivalence class of $x$ for $\sim_i$ ($i=1,2,3$).
Prove that $\sim_3$ is an equivalence relation on $A$. Prove that $[x]_3 = [x]_1 \cap [x]_2$.
I don't know if I can just conclude $\sim_3$ is an equivalence relation relation given that $\sim_1$ and $\sim_2$ are also equivalence relations on the same set. Do i still need to show the three properties and if so how would I do that for this problem?
For the second part I want to say that since $\sim_1$ and $\sim_2$ are disjoint relations on a $[x]_1$ may not equal $[x]_2$. Since $\sim_3$ is defined by $\sim_1$ and $\sim_2$, $[x]_3$ must include some elements of both which is achieved at the intersection of $[x]_1$ and $[x]_2$.
Any help would be great!
Yes, you have to prove that $\sim_3$ has the three defining properties of an equivalence relation. To prove that $\sim_3$ is reflexive, let $a\in A$. Then $a\sim_1 a$, since $\sim_1$ is reflexive, and $a\sim_2 a$, since $\sim_2$ is reflexive, so $a\sim_3 a$ by the definition of $\sim_3$. Since $a$ was an arbitrary element of $A$, this shows that $a\sim_3 a$ for each $a\in A$ and hence that $\sim_3$ is reflexive. Symmetry and transitivity of $\sim_3$ are proved in very similar fashion.
To show that $[x]_3=[x]_1\cap[x]_2$, show that each side is a subset of the other.
$[x]_3\subseteq[x]_1\cap[x]_2$: Let $y\in[x]_3$. Then by definition $y\sim_3x$. Now use this to show that $y\in[x]_1\cap[x]_2$; you’ll probably want to do this by showing that $y\in[x]_1$ and that $y\in[x]_2$.
$[x]_1\cap[x]_2\subseteq[x]_3$: Let $y\in[x]_1\cap[x]_2$. Then $y\in[x]_1$ so $y\sim_1x$. What else can you say about $y$?