A cap set in $\mathbb{F}_3^n$ is a subset that contains no three elements adding to 0. Building on a construction of Edel, a recent paper of Tyrrell gave the first improvement to the lower bound for a size of a cap set in two decades showing that, for large enough $n$, there is always a cap set in $\mathbb{F}_3^n$ of size at least $2.218^n$. This was shown by constructing what is called an $I(11,7)$ admissible set. An admissible set is a subset of $\{0,1,2\}^m$ such that the supports of the vectors form an antichain with respect to inclusion and each triple of vectors has some coordinate where either exactly one of them is non-zero or exactly two are and they have different values. Such an admissible set is said to be $I(m,w)$ if it is of size $\binom mw$ and all of the vectors have exactly $w$ non-zero elements. In Tyrrell's paper they conjectured that $I(m,w)$ admissible set exists for all parameters. We resolve this conjecture by showing that there exists an $N$ such that an $I(N,4)$ admissible set does not exist. We refer to the type of a vector in $\{0,1,2\}^m$ is the ordered sequence of its non-zero coefficients. The vectors of type $12$ form an $I(m,2)$ admissible set and the vectors of type $121$ form an $I(m,3)$ admissible set (as can be easily checked by an interested reader). Sadly it is quite easily proved that there is no $I(6,4)$ admissible set where all vectors are of the same type. It follows by Ramsey's Theorem applied to 4-regular hypergraphs that there exists an $N$ such that an $I(N,4)$ admissible set does not exist. A similar argument shows that there exists an $N'$ such that an $I(N',N'-2)$ admissible set does not exist. Since we can construct an $I(m-1,w)$ and an $I(m-1,w-1)$ admissible set from an $I(m,w)$ admissible set, it follows that there are only finitely many $I(m,w)$ admissible sets exist other than the known forms.