Let $R$ and $S$ be relations such that $R \subseteq S$. Prove that $R^n \subseteq S^n$ for all positive integers $n$. If $ R$ be a symmetric relation. Prove that $R^n$ is symmetric for all positive integers $n$.
This just seems strange since $R^{n+1} \subseteq S^{n+1}$ is just basically $R\cdot R^{n} \subseteq S\cdot S^{n}$, though this doesn't tell me much and don't know what to use. I also see the induction hypothesis is in what I wrote, but don't know how to use it either.
Remember that $R\subseteq S$ means that whenever $aRb$ (i.e. $(a,b)\in R$) it follows that $aSb$, (i.e. $(a,b)\in S$). Now, given two relations $R_1\subseteq A\times B,R_2\subseteq B\times C$ we define the composition $R_1R_2\subseteq A\times C$ as follows: we say that $aR_1R_2b$ iff there is a $z\in B$ such that $aR_1z$ and $zR_2b$ (shortly, $aR_1zR_2b$). Thus $aR^2b$ iff there exists $z$ such that $aRzRb$, and subsequently $aR^nb$ iff there exist $z_1,\ldots,z_n$ such that $aRz_1Rz_2R\cdots Rz_nRb$. But, remember that $aRz_1$ implies $aSz_1$, $z_1Rz_2$ implies $z_1Sz_2$ and so on. Ultimately, you get $aS^nb$, which is what you sought for.
If $R$ is reflexive, that is $aRa$ for each $a$, then we have that $aRa, aRa$ which gives $aR^2a$. So $R^2$ is reflexive. And so on: $aRaRaR\cdots Ra$ gives $aR^na$.
