What does the cut-off $\Lambda$ stand for in the theory of QED? The bare electron mass $m_0$, in QED, changes as $$m_0\to m=m_0+\delta m\Big(\frac{\Lambda}{E}\Big)$$ where high momentum modes from $E$ to $\Lambda$ has been integrated out. 
What scale does the cut-off $\Lambda$ stand for in the theory of QED and why?  Is it the top quark mass $\Lambda=m_{top}$, the GUT scale $\Lambda=M_{GUT}$ or the Planck scale $\Lambda=M_{Pl}$? I never understood which scale corresponds to the correct cut-off of a theory.
 A: There is usually no unique cutoff scale $\Lambda$ in renormalization. The reason is that generally we don't know what the ultimate microscopic physics is.
So the rationale is to pick any scale $\Lambda$ to be much, much larger than any physical scale of interest (particle mass or energy $E$ of an experiment you're doing) and then adjust couplings as you lower $\Lambda$ - the usual RG flow story.
In some cases you do get information about the range of validity of a theory. Say you have a theory with some fields $\Phi_i$ which couple to a heavy particle $X$ of mass $M$, and you integrate out the heavy particle. Then you obtain an effective action for the $\Phi_i$ fields which is valid up to energies $E < M$. This is reflected by the fact that you generate dimensionful couplings of size $M$ to the appropriate power. A physically interesting example is the chiral Lagrangian for pion physics:
$$L = \frac{f_\pi^2}{2} \text{Tr}(\partial_\mu \Sigma \partial_\mu \Sigma^\dagger) = \frac{1}{2} | \partial \vec{\pi}|^2 + \frac{1}{f_\pi^2}  \left[\vec{\pi}^2(\partial \vec{\pi})^2 - (\vec{\pi} \partial \vec{\pi})^2 \right] + \ldots
$$
All couplings scale like the pion decay constant $f_\pi$, and this action is useful to compute $\pi \pi$ scattering at low energies but breaks down below $\Lambda_\text{QCD}$. QED is however not of this form: it has a single, dimensionless coupling $\alpha$, which clearly doesn't carry any information about scales. Moreover, QED isn't a theory of quarks, gravity or weak interactions, so there's no way to tie $\Lambda$ to $m_\text{top}$, $M_\text{GUT}$ or $1/\ell_\text{pl}$. 
