The notion of a germ in singularity theory I quote from my lecture:

Let $X$ be a topological space (think of $X=\mathbb{C}^n$ with the classical topology), $p\in X$, $A,B\subseteq X$. Then $A\sim B$ if there exists an open subset $U\subseteq X$, $p\in U$, and $A\cap U=B\cap U$. The germ $(A,p)$ of $A$ at $p$ is the equivalence class of $A$ under this equivalence relation.

So far okay. After this, we simply wrote we'd from now on consider the case $X=V(I)\subseteq\mathbb{C}^n$, where $I$ is an ideal in $\mathbb{C}[x_1,...,x_n]$ or $\mathbb{C}[[x_1,...,x_n]]$, and look at $(X,p)$ instead of at $X$. The first place where the notion occurs again is this:

If $f$ is a polynomial (or a convergent power series) we may consider $(V(f),0)\subseteq(\mathbb{C}^n,0)$ and find $U\subseteq\mathbb{C}^n$ open, such that $f$ is defined on $U$, and $0$ is singular, but all $p\in U\smallsetminus\{0\}$ are non-singular. (This is written after a motivating example on isolated singularities, it is not possible for an arbitrary $f$ in general.)

While it is intuitively clear to me what an isolated singularity should be (geometrically), the term including the germs still confuses me there. If I didn't misinterpret the definition, $(\mathbb{C}^n,0)$ should consist of all neighbourhoods of $0$ (all sets $A$ in $\mathbb{C}^n$ s.t. there exists an open set $U$ with $0\in U$, and $A\cap U=U$, i.e., $U\subseteq A$). Why is then $(V(f),0)\subseteq(\mathbb{C}^n,0)$? In general, $V(f)$ contains no classical open neighbourhood of $0$, so why should any $A$ with $A\sim V(f)$ be in $(\mathbb{C}^n,0)$?
Thanks for your help in advance! I bet it's simple error I'm having again ;)
 A: Well no, $A\sim B$ if there exists a neighbourhood $U$ of $p$ such that $A\cap U = A\cap U$ but the set $U$ is not necessary contained in $A$ nor $U$ contans necessary $A$. I would also point out that every two neighbourhoods of $p$ are related by this equality relation. In fact their intersection is a neightbourhood of $p$.
The germ $(A,p)$ is, practically, the set of all the set that looks like $A$ sufficiently near the point. 
A: This has been 'answered' some time ago by me looking into the (hardly available) book my professor has written: Let $X$ be a topological space and $x\in X$. For two germs $(Y,x)$, $(Z,x)$ at $x$, we write $(Y,x)\subseteq(Z,x)$ if there exist representatives $Y$ of $(Y,x)$ and $Z$ of $(Z,x)$ with $Y\subseteq Z$. It is only a notation, but we didn't mention it in the lecture, hence I was a little confused.
It simply has the advantage that, while as sets the germs are not contained in each other, we still have the relation: $(Y,x)\subseteq(Z,x)$ and $(Z,x)\subseteq(Y,x)$ $\Rightarrow$ $(Y,x)=(Z,x)$.
