Trivial proof writing regardings reflexive relations

Question

Q: Suppose $R_{1}$ and $R_{2}$ are relations on A. Give a proof or counterexample to justify your answer. If $R_{1}$ and $R_{2}$ are reflexive, must $R_{1} \cup R_{2}$ be reflexive?

A: My reasoning is as follows:

Let $A$ be a set. Let $\alpha \in A$ and $\alpha$ is arbitrary. Let $R_{1}$ be a relation on $A$. Let $R_{2}$ be an arbitrary relation on A.

If $\alpha \in A$ such that $R_{1}=\{(_{\alpha}R_{\alpha})\}$ then $R_{1}$ is reflexive.

Then $R_{1} \cup R_{2} = \forall \alpha \in A((_{\alpha}R_{\alpha}))$

Since $\alpha$ were arbitrary, this proves that $R_{1} \cup R_{2}$ must be reflexive.

$\blacksquare$

Please let me know what you think and how I can improve my proof writing. Thank you.

Answer 1

I think you may have misunderstood something about the relation $R_1\cup R_2$, but I'm not completely sure what it is. Let me try to explain what the relation $R_1\cup R_2$ looks like, and then give my version of a proof why it is reflexive.

You are given that $R_1$ and $R_2$ are two reflexive relations. If $A=\{1,2,3\}$, then you might have that

$$R_1 = \{(1,1),(1,2),(2,2),(3,3)\}$$ and $$R_2 = \{(1,1),(2,1),(2,2),(3,1),(3,3)\}.$$ You see that these two relations are both reflexive, because for all $x\in A$, we have that $(x,x)\in R_1$, and also $(x,x)\in R_2$.

From these two relations, we make a new relation, namely the union $R_1\cup R_2$. This is just one new relation on $A$. Let me call it $R$, so $R=R_1\cup R_2$. What does $R$ look like? It is just the union of the relations $R_1$ and $R_2$, so if $(x,y)$ is a pair of either $R_1$ or $R_2$, then it will also be a pair of $R$. We have $$R = \{(1,1),(1,2),(2,1),(2,2),(3,1),(3,3)\}.$$ You see that this relation is still reflexive, so this is an affirmative example of what you are trying to prove.

Now to the general proof. Take $x\in A$. I want to prove that $R=R_1\cup R_2$ is reflexive. Since $R_1$ and $R_2$ are both reflexive by assumption (I really only need one of them to be reflexive), then for any $x\in A$, I have that $(x,x)\in R_1$ and $(x,x)\in R_2$. Since $R=R_1\cup R_2$, I then also have that $(x,x)\in R$. This proves that $R_1\cup R_2$ is reflexive.

Answer 2

$R_1, R_2$ are reflexive relations on the set $A$.

This means $\forall x \in A: (x,x)\in R_1$ and $\forall x\in A: (x,x)\in R_2$

We wish to investigate whether $R_1\cup R_2$ is (or is not) reflexive.

Well: $$\begin{array}{l|l:l} 1. & \forall x\in A: (x,x)\in R_1 & \textsf{Premise 1} \\ 2. & \forall x\in A: (x,x)\in R_2 & \textsf{Premise 2} \\ 3. & \forall x\in A:\Big( (x,x)\in R_1 \wedge (x,x)\in R_2\Big) & 1,2,\wedge\textsf{Introduction} \\ 4. & \forall x\in A :\Big((x,x)\in R_1\vee (x,x)\in R_2\Big) & 3, \textsf{Weakening} \\ 5. & \forall x\in A : (x,x)\in R_1\cup R_2 & 4, \textsf{Definition of Union} \\ \hline \Box & \Big(\big(\forall x\in A: (x,x)\in R_1\big)\wedge\big(\forall x\in A: (x,x)\in R_2\big)\Big)\to \big(\forall x\in A: (x,x)\in R_1\cup R_2\big) & 1,2,5,\to\textsf{Introduction} \end{array}$$

So if $\lower{0.25ex}R_1$ is reflexive and $\lower{0.25ex}R_2$ is reflexive, then their union $\lower{0.25ex}{R_1\cup R_2}$ is also reflexive.