Quote:
Originally Posted by MaciekLeks
Yes, I tried to write it in Context part of my post.
Unfortunately it does not help. I understand the assumption, but I do not understand the source of confidence that we can correlate with perceptron outputs.

I'm not sure if I understand your point right:
If the perceptron hypothesis set can shatter the data set then for the same
, there exists
and
such that
and
. It means that if the data set is shatterable by the perceptron hypothesis set, for any
in the data set, we can choose the case to be
or
and be sure that there exists at least one hypothesis from perceptron hypothesis set can implement the case we have chosen.
Instead of choosing
or
explicitly, the proof let the choice for the value of
depends on the value of
:
, because for whatever the real value of
is,
only has value of 1 or 1, hence
or
. In my understanding, this dependence does not make the chosen dichotomy invalid.