= Exclusion versus particle position
Exclusion constrains complete <one-particle states>, not a collection of little spatial boxes. In particular, two opposite-<spin> <Electrons> in the same <spatial orbital> have
$$
\Psi(\mathbf r_1,s_1;\mathbf r_2,s_2)=\phi_a(\mathbf r_1)\phi_a(\mathbf r_2)\frac{\chi_\uparrow(s_1)\chi_\downarrow(s_2)-\chi_\downarrow(s_1)\chi_\uparrow(s_2)}{\sqrt2}.
$$
The spatial factor is symmetric; the <spin-one-half singlet state> supplies the required antisymmetry. At $\mathbf r_1=\mathbf r_2$, this <wavefunction> need not vanish for opposite <spin> labels. For equal <spin> labels it does vanish. In a continuous position description, an exact coincident point has zero measure anyway; the useful comparison concerns joint <probability densities> or detection in finite regions. Equal positions do not imply equal complete <quantum states>.
The <fermionic occupation numbers> $n_\alpha=0,1$ refer to a chosen <orthonormal basis> of <one-particle states>. A <quantum superposition> need not have definite <occupation numbers> in that basis, and an <expectation value> $\langle n_\alpha\rangle$ can be fractional. For example, one <Electron> in $(u+v)/\sqrt2$ has mean <occupation number> $1/2$ in each of two <orthogonal> <spin orbitals>, without splitting into two particles. The <canonical anticommutation relations> give $(a_\alpha^\dagger)^2=0$ and make $a_\alpha^\dagger a_\alpha$ a <projector>, so its <expectation value> stays between zero and one in every normalized <quantum state>.
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