Let $R$ be a relation on a nonempty set $X$. Let $R_0=R$ and for every $n\in\mathbb{N}$, define the relation $R_n$ on $X$ by $xR_ny$ iff there exists $z_1,z_2,\dotsc,z_n\in X$ s.t. $xRz_1,z_1Rz_2,\dotsc,z_nRy$. The relation $tr(R)=R_0\cup R_1\cup\dotsb$ is called the transitive closure of $R$. Show that $tr(R)$ is transitive, and if $R'$ is a transitive relation with $R\subseteq R'$, then $tr(R)\subseteq R'$.
Can anyone check my proof?
Let $xtr(R)y,ytr(R)z$, then $xR_ny,yR_mz$ for some nonnegative integers $n,m$.
CASE 1: If $n,m=0$, then $xRy,yRz$ and therefore $xR_1z$ so $xtr(R)z$.
CASE 2: Suppose say $n\neq0$, then there exists $z_1,z_2,\dotsc,z_n\in X$ s.t. $xRz_1,z_1Rz_2,\dotsc,z_nRy$. Then for $z_1,z_2,\dotsc,z_n,y\in X$, $xRz_1,z_1Rz_2,\dotsc,z_nRy,yRz$ and therefore $xR_{n+1}z$ so $xtr(R)z$.
CASE 3: Similarly, if $n,m\neq0$, then for $z_1,z_2,\dotsc,z_n,y,\alpha_1,\alpha_2,\dotsc,\alpha_m\in X$, $xRz_1,z_1Rz_2,\dotsc,z_nRy,yRz,xR\alpha_1,\alpha_1R\alpha_2,\dotsc,\alpha_nRy$. It follows that $xR_{n+m+1}z$ and therefore $xtr(R)z$.
In every case $xtr(R)z$ so $tr(R)$ is transitive.
2nd part:
Let $xtr(R)y$, then $xR_ny$ for some nonnegative integer $n$. If $n=0$, then $xRy$ and therefore $xR'y$ so $tr(R)\subseteq R'$. If $n\neq0$, then there exists $z_1,z_2,\dotsc,z_n\in X$ s.t. $xRz_1,z_1Rz_2,\dotsc,z_nRy$. Since $R\subseteq R'$, $xR'z_1,z_1R'z_2,\dotsc,z_nR'y$, and since $R'$ is transitive, $xR'y$ so $tr(R)\subseteq R'$.
Your proof is the correct. You wrote it down clearly too. This is what my profs would call the 'follow your nose proof', it just rolls out of the definitions.
Since it is boring to just leave it at that, i'll nitpick. The expression $xtr(R)y$ is very hard to read, I would recommend that you give the operation a different formatting. For example $x\mbox{tr}(R)y$, where the transitive closure operation is formatted with \text{..}.