Let $A$ be a set and $R,S$ are relations on $A$. Then.... Let $A$ be a set and $R,S$ are relations on $A$. Define the relation $R+S$ on $A$ as $$R+S = (R-S)\cup(S-R)$$
Now show that

*

*If $A$ is countable then so does $R+S$

*Give and example of a set $A$ that is uncountable but $R+S$ is countable.

*Prove or disprove: If $R$ and $S$ are partial orders on A then so does $R+S$
For the first part, I am thinking that a relation gives us subset of the set on which it is defined, which is in our case $A$ so if $A$ is a countable set then its subset is countable as well.
For the second part, Lets say $A=\mathbb{R}$ relation $S$ gives us rational numbers (or its subset) and $R$ gives us Natural numbers (or its subset) then their symmetric difference which is $R+S$ becomes a countable set however $\mathbb{R}$ is uncountable.
For third part, I am thinking of relations $R=S=\{(a,a)|a\in \mathbb{Z} \}$ which holds both reflexive and transitive property but I am not sure about anti symmetry. Also this is just an example. How can we prove (if it is right) or disprove (Give counter example)
Please help
 A: A relation $R$ on a set $A$ is not a subset of $A$: it is a subset of $A\times A$. That means that your answer to the second question is wrong from the start: relations on $\Bbb R$ are not subsets of $\Bbb R$. You could, however, let $A=\Bbb R$ and let $R=S$ be any relation on $\Bbb R$, say $\{\langle x,x\rangle:x\in\Bbb R\}$, the identity relation on $\Bbb R$. Then
$$R+S=(R\setminus S)\cup(S\setminus R)=\varnothing\cup\varnothing=\varnothing\,,$$
which is certainly countable.

If the source of this problem misuses countable to mean countably infinite, you’ll have to work a little harder. You could, for instance, let $$R=\{\langle n,n+1\rangle:n\in\Bbb N\}$$ and $$S=\{\langle n,n-1\rangle:n\in\Bbb Z\}\,;$$ then $$R+S=(R\setminus S)\cup(S\setminus R)=R\cup S\,,$$ which is countably infinite.

Something very much like your idea, but with a correct understanding of what a relation on $A$ is, can be used to answer the first question. If $A$ is countable, then so is $A\times A$, and therefore so is every subset of $A\times A$, i.e., every relation on $A$.
Your example for the third question will work: your relation is a partial order, and $R+S=R+R=\varnothing$ is not reflexive, so it is not a partial order.
