What does the condition of $T_{3\frac{1}{2}}$ space mean exactly? 
A topological space $(X,\mathcal{T})$ is said to be $T_{3\frac{1}{2}}$
  if given $x \in X$, and a closed set $C \subset X$, $x \not \in C,
 \exists f:X \to [0,1]$ s.t. $f(x) = 0, f(C) = 1$

This condition seems to be a little bit arbitrary, what is the implication for a function to map a point and closed set to $0,1$?
What is an example of such a space?
 A: The Tychonoff spaces are the spaces with a special relation to the reals: the are precisely the spaces we can embed into (so can see as esentially subspaces of) $\mathbb{R}^I$ for some set $I$.
They are precisely the spaces $X$ that have a compactification (a space $Y$ in which we embed $X$ as a dense subset and such that $Y$ is compact Hausdorff).
They are the natural spaces to study the ring of (real-valued) continuous functions $C(X)$ on. Etc. 
So these spaces popped up so naturally, so we needed a name for it, and it's inbetween $T_3$ and $T_4$ (the latter by Urysohn's lemma, which let's us separate two closed disjoint sets in a normal space not just by open sets, but even by a continuous function). 
A: $T_{3\frac{1}{2}}$ spaces are also called Tychonoff spaces or completely regular. They fall between $T_3$ and $T_4$ spaces with respect to the separation axioms.  In particular, Urysohn's lemma shows that every $T_4$ space is completely regular.
It is an exercise in Folland's Real Analyis to show the following:
If $(X,\mathcal{T})$ is completely regular, then $\mathcal{T}$ is the weak topology generated by $C(X)$. 
The Moore plane is an example that is $T_{3\frac{1}{2}}$ but not $T_4$.
