Suppose $G_1 = (V_1,E_1)$ and $G_2 = (V_2,E_2)$ be two graphs, then union of two graphs is $G_1\cup G_2 = (V_1\cup V_2, E_1\cup E_2)$ and $G_1\cap G_2 = (V_1\cap V_2, E_1\cap E_2)$. Now, it is given that $G_1$ is isomorphic to $G_2$, denoted as $G_1\simeq G_2$.
My question is (very basic question, not found in any graph theory text book),
Suppose we consider two different graph $G_1^{'}= (V_1^{'},E_1^{'})$ and $G_2^{'}= (V_2^{'},E_2^{'})$ and $G_1\simeq G_1^{'}$ and $G_2\simeq G_2^{'}$. Then can I say that,
- $G_1\cup G_2 \simeq G_1^{'}\cup G_2^{'}$,
- $G_1\cap G_2 \simeq G_1^{'}\cap G_2^{'}$.
If not, is there any counter example.
Small Modification
I make a small assumption, that is at which position of the graph, common vertex appear that does not change for $G_i^{‘}$ for $i=1,2$ also. Does this assumption helps in preserving the set operation?