$M$ is a non-empty set and $R \subseteq M \times M$ is a relation on $M$. A R-Clique is a set $C \subseteq M$, with for every $x$ and $y$ in $C$: $xRy$.
a)Prove or disprove: If $|M| > 1$ and R is reflexive and symmetrical, then there exists an R-Clique $C$ with at least two elements($|C| \geq2$).
b)Now let there be two R-Cliques $C_1,C_2$ with $M = C_1 \cup C_2$ and $C_1 \cap C_2 = \emptyset$. Prove or disprove: R is an equivalence relation (reflexive, symmetrical and transitive).
a) (i)$|M| > 1$
(ii)$\forall x,y \in M: xRx$ and $xRy \rightarrow yRx$
To prove: $i$ and $ii \rightarrow |C| \geq 2$
Since $C \subseteq M$, $C$ can also be entire $M$, which means $C=M$. Since $|M| > 1 \rightarrow |C| > 1 \rightarrow |C| \geq 2$.
b)I have to show that $R$ is an equivalence relation on $C_1$ and on $C_2$. Then it will also be an equivalence relation on M. But how do I show that since it's not specified what $R$ actually does.
For (a), you say
It's true that that is a possibility, but it's hardly a necessity that $C=M$. (In fact, the claim they ask you to prove is not true. See if you can think of a symmetric and reflexive relation on $M$ where all of its cliques are of size 1.
For (b), they are defining $xRy$ as it either being the case that $x\in C_1$ and $y\in C_1$ or that $x\in C_2$ and $y\in C_2$. This is enough of a definition for you to determine that $R$ is an equivalence relation.