On Goldstone fermions / goldstino: SUSY breaking

There are statements about Goldstone fermions, or goldstino, seem confusing to me.

(1) Goldstone boson requires a continuous symmetry spontaneously broken. Does Goldstone fermion imply continuous SUSY spontaneously broken? But boson B and fermion F Hilbert space can be finite dimensional and discrete. The number of SUSY charge $mathcal{N}$ is also discrete. What do we mean by continuous SUSY if there is ? If not, how does the Goldstone theorem on the continuous global symmetry applies to SUSY (possibly discrete, switching between bosonic B and fermionic F sector, and finite $mathcal{N}$)?

(2) In Witten 1982 NPB – Constraints on Supersymmetry Breaking, he said:

(A) the number of boson zero energy modes $n_b=0$ and fermion zero energy modes $n_f=0$ — supersymmetry is spontaneously broken and there is a massless Goldstone fermion in the infinite volume theory.

• How and why the massless Goldstone fermion appears, only in the infinite volume limit? (only when the states of bosonic B and fermionic F sector become infinite?) How to interpret Goldstone fermion when we have a finite size but large $L$ system?

(B) the number of boson vs fermion zero energy modes $n_b= n_fneq 0$, supersymmetry is not broken; there is no Goldstone fermion, but there are zero-energy fermionic states which we interpret as evidence that the infinite volume theory has a massless fermion (a massless fermion is not a Goldstone fermion unless it is created from the vacuum by the supersymmetry current).

• How do we know there is a massless fermion (in the infinite volume)?

• How to show Goldstone fermion is created from the vacuum by the supersymmetry current? (in the infinite volume? How to interpret it in the finite volume?)