Equivalence relations on a set and partial order

Question

I am new here and this is the first time that I am asking a question here, and it's also the first time that I am using $\LaTeX$, so I apologize if I make some mistakes. I study in a non-English university, so I am sorry if my questions are a bit flimsy. Due to universities being closed, I have to learn everything from home with some class notes that my teacher put online. I am taking a discrete mathematics class for the first time, and I find it do-able, but it gets a bit hard to study it by myself. I have been doing some exercises regarding relations, but I ran into some difficulties for these two questions:

1) Let $R$ be a relation defined on $\mathbb{R}_+ \times \mathbb{R}_+$ where $(x_1,y_1)\mathrel{R}(x_2,y_2)$ if and only if $x_1 \times y_2 \leq x_2 \times y_1$. Prove or refute the following : $R$ is a partial order.

2) Let $R_1$ and $R_2$ be two relations defined on $X$. Show that if $R_1$ and $R_2$ are equivalence relations, then $R_1 \cap R_2$ defined on $X$ is also a equivalent relation.

Here is what I have been able to do so far:

1) I don't know how to do this. I really need help with this kind of question.

2) To show that $R_1 \cap R_2$ is an equivalence relation when $R_1$ and $R_2$ are equivalence relations, I supposed that if $x\mathrel{(R_1 \cap R_2)}y$ and $y\mathrel{(R_1 \cap R_2)}z$, then $x \mathrel{R_1} y$ and $y \mathrel{R_1} z$, so $x \mathrel{R_1} z$ and for $R_2$ we have $x \mathrel{(R_1 \cap R_2)} z$. I think that this proves that $R_1 \cap R_2$ is transitive, but I am not sure. Assuming this is correct (please explain how), how could I phrase this in a proper way?

Thanks! Your help is really appreciated.

Answer 1

1) To see whether this relation is a partial order or not, you need to check the three properties in the definition. For example, to show reflexivity, take $(x,y) \in \mathbb{R}_+ \times \mathbb{R}_+$ and check if $(x,y)R(x,y)$ follows. Similarly, for antisymmetry take two pairs $(x_1, y_1), (x_2, y_2)$, then assuming $(x_1, y_1)R(x_2, y_2)$ and $(x_2, y_2)R(x_1, y_1)$, use the definition of $R$ to see if these pairs have to be equal. Follow the same procedure with three pairs for transitivity. Once you plug in what $R$ means into your assumptions, these properties follow or fail shortly.

2) Your reasoning is correct. Start by stating your hypotheses clearly: "Let $x(R_1 \cap R_2)y$ and $y(R_1 \cap R_2)z$". I sometimes like to say "We want to show that $x(R_1 \cap R_2)z$", but that's a personal preference. Then, write down your reasoning in enough detail. For example, depending on the expected level of detail you might need to say "... then $xR_1z$ by the transitivity of $R_1$" or keep it brief like you did. Keep doing this for the three properties of equivalence relations and finally say that this completes the proof of $(R_1 \cap R_2)$ being an equivalence relation.

Answer 2

Hint for (1). Recall that a partial order is a binary relation that is reflexive, antisymmetric and transitive. Thus you need to prove or disprove (by giving a counterexample) that the following properties hold for all $x_1, x_2, y_1, y_2$ in ${\Bbb R}_+$.

1. (reflexivity) $x_1y_1 \leqslant x_1y_1$
2. (antisymetry) if $x_1y_2 \leqslant x_2y_1$ and $x_2y_1 \leqslant x_1y_2$, then $x_1 = x_2$ and $y_1 = y_2$.
3. (transitivity) if $x_1y_2 \leqslant x_2y_1$ and $x_2y_3 \leqslant x_3y_2$, then $x_1y_3 \leqslant x_3y_1$.

If one of these three properties fails to be true, then your relation is not a preorder.

(2) Your argument for transitivity is correct. You could start by saying

By definition, $$ (*)\quad x \mathrel{(R_1 \cap R_2)} y \iff x \mathrel{R_1} y \ \text{ and }\ x \mathrel{R_2} y. $$ and then use $(*)$ to prove that $R_1 \cap R_2$ is reflexive, symmetric and transitive.