Union and intersection of two transitive relations

If we have two transitive relations $$R_1, R_2$$ on a set $$A$$, then I know that if $$R_3 = R_1 \cup R_2$$ that $$R_3$$ is not transitive.

And if $$R_3 = R_1 \cap R_2$$ then $$R_3$$ is transitive. But I can't figure out why my "proof" stating the opposite of those is false.

My reasoning went as follows: For $$R_3 = R_1 \cup R_2$$, Let $$x,y,z\in A$$ and assume $$(x,y),(y,z),(x,z)\in R_1$$. Since $$R_1\subseteq R_1 \cup R_2 = R_3$$ then $$(x,y),(y,z),(x,z)\in R_3$$ and similarly the same argument for $$R_2$$ so $$R_3$$ is transitive.

For $$R_3 = R_1 \cap R_2$$ let $$A = \lbrace 1,2,3,4 \rbrace$$ and have $$R_1 = \lbrace (1,2),(2,3),(1,3) \rbrace$$ and $$R_2 = \lbrace (1,2),(2,4),(1,4) \rbrace$$ hence $$R_1$$ and $$R_2$$ are transitive but $$R_3 = \lbrace (1,2) \rbrace$$ which is not transitive

Can someone point out my logical flaws in both arguments? I know one can show counter examples for the first one but I don't see where I'm making poor arguments in both of them as I was told we can handle relations just like any other type of set.

Firstly, in order to prove transitivity of some relation $$R$$, you need to show that $$(x,y),(y,z)\in R$$ implies $$(x,z)\in R$$.

The problem with your "proof" for the union is that you assume that $$(x,y),(y,z)$$ both belong to the same relation $$R_i$$. In that case, you indeed have $$(x,z)\in R_i$$ and hence $$(x,z)\in R_3$$. The problem occurs when you take $$(x,y)\in R_1$$ and $$(y,z)\in R_2$$. For example, if $$R_1=\{(x,y)\in\mathbb N^2\mid x-y\text{ is divisible by 2}\}$$ and $$R_2=\{(x,y)\in\mathbb N^2\mid x-y\text{ is divisible by 3}\}$$, both $$R_i$$ are transitive, but $$R_1\cup R_2$$ isn't: Consider $$(0,2)\in R_1$$ and $$(2,5)\in R_2$$. Do you think $$(0,5)\in R_1\cup R_2$$?

Your "counterexample" concerning the intersection isn't actually a counterexample! $$R_3$$ is indeed transitive. More precisely, we need to prove $$\forall x,y,z: (x,y),(y,z)\in R_3 \implies (x,z)\in R_3.$$

This implication is vacuously tue, that is, the antecedent (the part before $$\implies$$) is never satisfied, and thus the implication holds (here we use that the implications "false implies true" and "false implies false" are both correct).

This is really two questions:

Why $$R_3 = R_1 \cup R_2$$ is not transitive. The property must hold for two arbitrary elements of the relation. Say $$(x, y) \in R_1$$ and $$(y, z) \in R_2$$ but not the reverse. They are both elements of the relation $$R_3$$ but there's no guarantee that $$(x, z) \in R_3$$ inherited from either $$R_1$$ or $$R_2$$. You can come up with a specific small counterexample.

Why your example doesn't contradict the fact that $$R_3 = R_1 \cap R_2$$ is transitive. Basically, there are no pairs of the form $$(a, b), (b, c) \in R_3$$, so the property holds vacuously.