Confusion of the definition of a completely regular polygon P I am a student studying an algebra & geometry module with an exam at the end of January. 
I have noticed that there are two alternate definitions of a completely regular polygon P:
1) A topological space X such that for every closed subset C of X and every point  x in X\C, there is a continuous function f:X->[0,1] such that f(x)=0 and f(C)={1}.
2) A topological space XX is said to be completely regular space, if every closed set AA in XX and a point x∈Xx∈X, x∉Ax∉A, then there exist a continuous function f:X→[0,1]f:X→[0,1], such that f(x)=0f(x)=0 and f(A)={1}f(A)={1}.
Now my question is, why are these two definitions equivalent? 
 A: The trick is that the two "inverses" in the two definitions are not the same thing.
As you noticed, the latter "inverse" satisfies always definition 1).
However the converse does not always hold.
But if there is such an $x$ as in definition 1), there exists always another element which satisfies definition 2).
This element is in fact $a^{-1}:= xax$ .
(If you go further, you can also notice that such an element must be unique to $a$.)
A: There is actually a third (and in my opinion, more suitable) definition of a completely regular semigroup: a completely regular semigroup is a semigroup in which every element is in some subgroup of the semigroup. With this definition in mind, your question becomes easy: you have to look for the group containing $a$.
It remains to prove that (1) implies (2). Let $e = ax = xa$. Then $e$ is idempotent since $ee = axax = ax = e$. Moreover, $a \mathrel{\mathcal{R}} e$ since $ax = e$ and $ea = axa = a$. In the same way, $a \mathrel{\mathcal{L}} e$ since $xa = e$ and $ae = axa = a$. Thus $a$ belongs to the $\mathcal{H}$-class of the idempotent $e$, which is, as you certainly know, a group, say $G$. Let $a^{-1}$ be the inverse of $a$ in $G$. Then the properties (i), (ii) and (iii) are trivially satisfied.
