Suppose $R$ is a reflexive and symmetric relation on a finite set $A$. Define a relation $S$ on $A$ by declaring $xSy$ if and only if for some $n∈ℕ$ there are elements $x_1,x_2,…,x_n∈A$ satisfying $xRx_1,x_1Rx_2,x_2Rx_3,x_3,Rx_4,x_{n−1}Rx_n$ and lastly $x_nRy$.
Prove $S$ is an equivalence relation and $R \subseteq S$.
Edit: I didn't include what my thoughts are or what I have done so far, so here it is: I know I need to prove $S$ is an equivalence relation and for that I need to show it is reflexive, symmetric and transitive. I first started by showing it was reflexive:
$xSx$: $xSx_1, x_1Sx_2,...,x_{n-1}Sx_n, x_nSx$. My reasoning for this was if I just set $y=x$, then the above holds. Although I am not confident about this reasoning.
Second, I tried to show $S$ is reflexive. In short, if $xSy$, then $ySx$. This is where I got really stuck as although I know $R$ is symmetric, I am not sure how to use that fact to show $ySx$. That is, $yRx_1,x_1Rx_2,...,x_{n-1}Rx_n,x_nRy$.
The middle ones make sense, that is, $x_1Rx_2$ is the same as $x_2Rx_1$, because $R$ is reflexive, but $yRx_1$ and $x_nRy$ don't seem to make sense to me.
All proofs are straightforward.
The relation $S$ is reflexive, because for each $x\in A$ there exist $n=1\in\Bbb N$ and $x_1=x\in A$ such that $xRx_1$ and $x_1Rx$ (each of the last expressions mean $xRx$, which is true by reflexivity of $R$).
The relation $S$ is symmetric. Indeed, let $xSy$. Then for some $n\in\Bbb N$ there are elements $x_1,x_2,...,x_n\in A$ satisfying $xRx_1$, $x_1Rx_2$,... $x_{n-1}Rx_n$ and lastly $x_nRy$. The symmetry of $R$ implies that $yRx_n$, $x_nRx_{n-1}$, ..., $x_2Rx_1$, $x_1Rx$, so $ySx$.
The relation $S$ is transitive. Indeed, let $xSy$ and $ySz$. Then for some $n,m\in\Bbb N$ there are elements $x_1,x_2,\dots,x_n\in A$ and $y_1,\dots, y_m\in A$ satisfying $xRx_1$, $x_1Rx_2$,... $x_{n-1}Rx_n$ and lastly $x_nRy$ and $yRy_1$, $y_1Ry_2$,... $y_{n-1}Ry_n$ and lastly $y_nRz$. For each $i$ from $1$ to $n$ put $z_i=x_i$ and for each $i$ from $n+1$ to $n+m$ put $z_i=y_{i-n}$. Then $xRz_1$, $z_1Rz_2$,... $z_{n-1}Rz_n$, $z_nRz_{n+1}$, $z_{n+1}Rz_{n+2}$,..., and lastly $z_{n+m-1}Ry$, so $xSz$.
$R\subset S$. Indeed, if $xRy$ then there exist $n=1\in\Bbb N$ and $x_1=x\in A$ such that $xRx_1$ (by reflexivity of $R$) and $x_1Ry$ (by $xRy$), so $xSy$.