Is the graph of the following type possible? Let a simple undirected graph be denoted by $(V, E)$, where $V$ is the set of vertices and $E$ is the set of edges. Let each vertex of a graph be a natural number. Now, I want to define the graph's edge such that $(V_1\cup V_2, E)=(V_1, E)\mathbin{\pmb\cup} (V_2, E)$, where $\cup$ and $\pmb\cup$ are set union and graph union, respectively. How should I define the graph's edge such that the edge is preserved (or, above equality holds)?
The following is my failed attempt.

let the edge of each graph represents "divides", i. e.,  vertices $ a$ and $b$ are  connected by an edge if and only if $a$ divides $b$ or, $b$ divides $a$. In this case, it is found that the above equality doesn't hold.

 A: Using the formulation from the discussion in the comments above, I don't think this is possible unless $f(X)$ only has edges between nodes and themselves for all $X$ i.e. $\forall X, f(X)\subseteq \{\{a,a\}|a \in X\}$.
Here is why: assume $X=\{n_1, ..., n_k\}$ and $f(X)=\{\{n_a,n_b\}, ...\}$ where $n_a \neq n_b$. Now, let's create two sets, $X_a=X-\{n_b\}$ and $X_b=X-\{n_a\}$. Then, we know:

*

*$\{n_a,n_b\} \notin f(X_a)$ since $n_b \notin X_a$

*$\{n_a,n_b\} \notin f(X_b)$ since $n_a \notin X_b$
But

*

*$\{n_a,n_b\} \in f(X)$

*$X=X_a \cup X_b$
To put it in graph terms, take a graph $G(E,V)$ and take any edge ${a,b}$. Then you can create two new graphs which don't have that edge by $a$ into one graph and $b$ into the other graph.
In fact, we can specify $f$ even more. The only way to get the desired property is that for any $a \in X$, $\{a.a\} \in f(X)$ iff $\{a.a\} \in f(\{a\})$ i.e. the question of whether or not to include $\{a,a\}$ is independent of the other vertices in $X$. (You can prove this using a similar argument to the above).
