The expression $P(A|B)$, where $(\Omega, \mathcal{A}, P)$ is a probability space and $A, B \in \mathcal{A}$ is only defined for when $P(B)>0$.
Suppose we have the following case: $([0,1], \mathcal{B} \cap [0,1], P)$, where $P(B)$ is the usual length of the measurable set $B$. Intuitively, we are picking a random real from $[0,1]$, where the likehood of it coming from any partition into equal parts is the same. So, it would make sense intuitively if $P(\{1\} | \{0,1\}) = \frac{1}{2}$. Is there any natural way to extend the concept conditional probability that agrees with this?
Measure theory teaches you that probabilities are in fact (generally, Lebesgue) measures (which means length in 1D, area in 2D, volume in 3D, n-hypervolume in nD). Your encompassing space $\Omega$ has measure (probability) 1. The probability of A knowing that B boils down to "what would the measure of A be, if B was a set of measure 1, instead of some fraction of $\Omega$ ?". It's a way of re-setting the problem so that B is the encompassing space.
There is no way to "scale" (or "ratio", I don't know which is the better term) a space of measure $0$ to make it into a space of measure $1$. That would require solving the equation $0x = 1$.
Of course, all of this depends on your measure. If your singleton set has a nonzero measure, nothing's preventing you from having fun with it. Then again, this means that $P(B) \neq 0$