Prove that $R\cap S$ is symmetric, transitive, and anti-symmetric.

Question

If you can confirm these are done correctly or offer another way to do so I would greatly appreciate it. Also how would you go about proving $R\cap S$ is reflexive? What assumption if any would be necessary?

Let $R$ and $S$ be relations on set $X$.

a) Assume $R$ and $S$ are symmetric.

Prove that $R\cap S$ is symmetric.

I believe the way to do so is:

Let $(x,y)\in R\cap S$

$\Rightarrow (y,x)\in R$ and $(y,x)\in S$ since R and S are symmetric

$\Rightarrow (y,x)\in R\cap S$

b) Assume $R$ and $S$ are transitive.

Prove that $R\cap S$ is transitive.

I believe the way to do so is:

Let $(x,y)\in R\cap S$ and $(y,z)\in R\cap S$

$\Rightarrow (x,y)\in R$ and $(x,y)\in S$

$\Rightarrow (y,z)\in R$ and $(y,z)\in S$

Then because $R$ and $S$ are transitive $(x,z)\in R$ and $(x,z)\in S$

$\Rightarrow (x,z)\in R\cap S$

c) Assume $R$ is anti-symmetric.

Prove that $R\cap S$ is anti-symmetric.

If $R$ is anti-symmetric any subset of $R$ is also anti-symmetric, also $R\cap S$.

If $(a,b),(b,a)\in R\cap S$

$\Rightarrow (a,b),(b,a)\in R$

$\Rightarrow a=b$

Answer 1

Your proof is correct. For reflexivity, note the following:

If $R$ and $S$ are reflexive relations on a set $X$ and $x\in X$. Then, $(x,x)\in R$ and $(x,x)\in S$ so that $(x,x)\in R\cap S$. I.e., $R\cap S$ is reflexive.