### Theorem:

An n-set Venn diagram cannot be created with circles for $n \geq 4$.

*Proof:*
Suppose it is possible.
Then we may create a planar graph $G$ from the Venn diagram by placing vertices at each intersection, like this:

$\small\text{(an example for a 3-set Venn diagram)}$

We will show that a graph created in such a way from a 4-set Venn diagram cannot be planar—a contradiction.

**Claim:** Every pair of circles intersect at two distinct points.

Suppose not. Then there is a pair of circles whose interiors do not intersect. This cannot be, for we require a Venn diagram graph to have a face for every pairwise intersection of sets.

**Claim:** Every vertex has degree four.

Suppose not. Then there are three circles $C_1, C_2, C_3$ which share an intersection. It can be readily checked by case analysis that not all intersections $\{C_1\cap C_2, C_1\cap C_3, C_2 \cap C_3, C_1\cap C_2 \cap C_3, \}$ are present as faces in this subgraph.

These two claims entail there are exactly two unique vertices for each pair of circles, so $v = 2{n \choose 2}$.
Also, by the edge sub formula we have $e=\frac{d_1+\cdots+d_v}{2}=2v=4{n\choose 2}$ edges. Because $D$ represents all possible intersections of $n$ sets, we have a face for each set alone, each pair of sets, each triplet of sets, and so on. Counting the outer face of the graph as ${n\choose 0}$, an application of the binomial theorem yields $$f= {n\choose 0} + {n\choose 1} + {n\choose 2} + \cdots + {n\choose n} = 2^n \text{ faces.}$$

We now substitute each of these results into Euler's formula:
$$v-e+f = 2{n\choose2}-4{n\choose 2} + 2^n= 2^n- 2{n\choose 2}\leq2,$$
which results in a clear contradiction for $n = 4$. To reach one for $n>4$, notice that if there existed such a Venn diagram for some $n > 4$ we could remove circles from the diagram to obtain one for $n=4$.

The following is a good resource for exploring the combinatorics of Venn Diagrams more deeply: http://www.combinatorics.org/files/Surveys/ds5/VennEJC.html