Can I bound the cardinality of a set if testing for membership in it is known to be NP-complete? I would like to have a bound on the cardinality of the set of unit disk graphs with $N$ vertices. It is known that checking whether a graph is a member of this set is NP-hard. Does this lead to any lower bound on the cardinality, assuming P $\neq$ NP? 

For example, suppose there is an ordering on all graphs with $N$ vertices. Would NP-hardness then imply the cardinality exceeds $2^N$, in that otherwise you could test for membership in polynomial time by doing a binary search through the set? I think this would assume that you have somehow stored the set in memory... Is this allowed? 

Defintion: A graph is a unit disk graph if each vertex can be associated with a unit disk in the plane, such that vertices are connected whenever their disks intersect.

Here is a reference on NP-hardness of membership testing for unit disk graphs:
http://disco.ethz.ch/members/pascal/refs/pos_1998_breu.pdf
Mahaney’s Theorem states that sparse NP-complete sets exist iff P=NP. Therefore, assuming $P \neq NP $ implies a super-polynomial lower bound on the number of instances of size $n$ in $NP$-complete sets, for infinitely many $n$. That is, if $P \neq NP$, then any $NP$-complete set must have some $\epsilon \gt 0$ such that for infinitely-many integers $n\ge 0$, the set contains 
at least $2^{n^{\epsilon}}$ strings of  length $n$. 

H. Buhrman and J. M. Hitchcock proved the lower bound ($2^{n^{\epsilon}}$) is tight, unless the polynomial hierarchy collapses. 

[1] H. Buhrman and J. M. Hitchcock, NP-Hard Sets are Exponentially Dense Unless coNP ⊆ NP/poly, In IEEE
Conference on Computational Complexity, pages 1–7, 2008

[2] Eric Allender, A Status Report on the P Versus NP Question, Advances in Computers, Volume 77, 2009, Pages 117-147