Proving that $s$ is a reflexive closure of $r$ I have this problem that I cannot figure out. Could you please help me out with this?
Let $r$ be a relation on the set $X$ and let $$R:=\{t:t \text{ is a  reflexive relation on } X \text{ with } r\subseteq t \}.$$ For $x,y \in X$ we have $x \, s \, y$ iff for each $t\in R$ we have $x \, t \, y$. Show that $s$ is the reflexive closure of $r$.
My idea so far was as follows.
Suppose $x,y \in X$ and $x \ s \ y$. It follows that for all $t\in R$ we have $x \ t \ y$. Since $r\subseteq t$ for all $t\in R$, it follows $r \subseteq s$ and also $s \subseteq t$. Since each $t\in R$ is reflexive, $s$ is also reflexive.
Could you please verify my idea? I'm sure it has some mistakes.
 A: Your reasoning makes no sense.
Start with an arbitary $x\in X$ and observe that $xtx$ is true for every $t\in R$. 
This because every $t\in R$ is reflexive. 
Then by definition we have $xsx$.
Proved is now that $s$ is reflexive.

For proving that $s$ is the reflexive closure of $r$ it is enough now to show that $r\subseteq s\subseteq t$ for every $t\in R$.
An interesting fact: the reflexive closure of $r$ is the set $r\cup\Delta$ where $\Delta:=\{\langle x,x\rangle\mid x\in X\}$. 
This set $\Delta$ is the so-called diagonal of $X$.
A: The reflexive closure of $r$ is the least relation $r'$ such that whenever $x \, r \, y$  and $x, y \in X$, we have $x \, r' \, x$ and $y \, r' y$ and $x \, r' \, y$. (From this definition, we immediately see that $r'$ is reflexive.)
We first show that $r' \subseteq s$ and then show that $s \subseteq r'$.
Suppose $x \, r' \, y$ and $x,y \in X$. 
There are now two cases. If $x = y$, then note that for every reflexive $t$ with $r \subseteq t$ we must have $x \, t \, x$. So $x \, s \, x$. If $x \neq y$, then we again have that $x \, t \, y$ for every reflexive $t$ with $r \subseteq t$. Again, $x \, s \, y$.
Next, suppose $x \, s \, y$. Again, there are two cases. If $x = y$, then by definition, we have $x \, r' \, y$. If $x \neq y$, then for every reflexive $t$ with $r \subseteq t$ we must have $x \, t \, y$. But this then means that $x \, r \, y$.
A: Yeah, this has some problems.

Since $r\subseteq t$ for all $t\in R$, it follows $r \subseteq s$ and also $s \subseteq t$.

How does it follow that $r \subseteq s$? That's not clear.
Fortunately, to show that $s$ is reflexive we don't need it ... yes, we need that $s$ is reflexive, but given its definition that's the easy part. 
The harder part is that at some point you  need to show that $s$ is the closure of $r$, i.e. that $s$ is the smallest reflexive relation for which $r \subseteq s$. That part you haven't addressed at all yet.
