This is a question motivated by elementary many-body quantum physics, thus I will start by giving some somewhat imprecise descriptions of the objects involved (I will also ignore spin, which is of no concern for this problem). For non-interacting systems, the many-body time-independent wave function in the position representation is of the form $\Phi(x_1,\ldots,x_n)=\text{Det}\left( \phi_j(x_k) \right)_{j,k=1}^{n}$ (here in this notation $\text{Det}\left( A(j,k) \right)_{j,k=1}^{n}$ means the determinant of the $n\times n$ matrix $A$ of elements $A(j,k)$ ), where each $x_j \in \mathbb{R}^3$ and the $\phi_j(x)$ are the so-called one-electron wavefunctions (they are the molecular orbitals). These determinants are known as Slater determinants. However, for interacting systems this is no longer true, although the wave function $\Psi(x_1,\ldots,x_n)$ is always antisymmetrical, i.e $\Psi(x_{\sigma(1)},\ldots,x_{\sigma(n)})=(-1)^{\vert \sigma \vert}\Psi(x_1,\ldots,x_n),$ for any permutation $\sigma \in S_{n},$ because of Pauli's exclusion principle for fermions.
A major topic in many-body physics is how far can one go approximating interacting wavefunctions by Slater determinants (Hartree-Fock or Density Functional Theory are examples of this idea), and in some sophisticated Quantum Chemistry methods, the idea is to employ linear combinations of Slater detrminants. Therefore, the question seems natural: Is the span of the set of Slater determinants "dense" in the subspace of antisymmetric functions? I use quotation marks because of course I haven't specified any particular functional space. Ideally, such a result would hold for a Sobolev space such as $W_{1,2},$ which I believe is the simplest space suitable for a mathematical treatment of the Schrödinger equation.
When we construct the space of antisymmetric functions, a formal way to do so is taking the so called exterior algebra. In the quantum mechanics case, assuming $\mathscr{H}_j$ is the one-particle Hilbert space of particle number $j$, we could take the $N$-fold wedge product
$$ \wedge_{i=1}^N \mathscr{H}_j$$
as our total Hilbert space. This is actually defined such that slater determinants are dense, because it is defined to be the closure of the following set:
$$ \mathrm{span} \{ f_1 \wedge f_2 \wedge \dots \wedge f_N | f_j \in \mathscr{H}_j \} $$
and this is just another way to write down the Slater determinant built out of $f_1,f_2,...f_N$.
So briefly: The antisymmetric wave functions are exactly the closure of the span of Slater determinants.