Holomorphic functions at $\Bbb C$ \ {$z_0$} The way i was taught, a function is holomorphic  in $z_0$ if it is (complex) differentiable in a neighborhood  of that point. However, holomorphic functions in C except $z_0$ are confusing me. Shouldn't the function be also holomorphic at $z_0$ if it is at all points on the neighborhood of $z_0$ ? What's incorrect on my thinking?
 A: Holomorphic function on $\mathbb C \setminus \{z_0\}$ are functions which are (complex) differentiable in every point of  $\mathbb C \setminus \{z_0\}$ 
A: Then the answer is rather topological and the key point is "differentiable in a neighbourhood of that point" or "some neighbourhood". 
Let's take any $z^{*} \in \mathbb{C} \setminus \{z_0\}$ Now, if we consider  $B_r(z^{*}, r)=\left\{  z \in \mathbb{C} | |z-z^{*}|<r \right\}$, i.e. centred in $z^{*}$ with radius $r=\frac{|z^{*}-z_0|}{2}$, this is an open disk containing $z^{*}$ and not containing $z_0$, which qualifies as "some neighbourhood".
Then, re "shouldn't the function be also holomorphic at $z_0$", 


*

*first of all, the funaction may not be defined at $z_0$. Or

*as trivial as the function not being derivable at $z_0$, like there exists at least 2 sequences $\{z_n\}$ and $\{w_n\}$ such that $$\lim_{z_n \rightarrow z_0} \frac{f(z_n)-f(z_0)}{z_n-z_0} \ne \lim_{w_n \rightarrow z_0} \frac{f(w_n)-f(z_0)}{w_n-z_0}$$

A: 
Shouldn't the function be also holomorphic at $z_0$ if it is at all points on the neighborhood of $z_0$?  

No, and for the following reason: the sizes of the neighbourhoods are not chosen in advance. Every point is differentiable on its own neighbourhood, and the size of that neighbourhood depends on the point we choose. In particular, since $\mathbb{C} \setminus \{z_{0}\}$ is open, we can always choose these neighbourhoods to be sufficiently small to exclude the point $z_{0}$.  
For instance, $z \to \frac{1}{z}$ is holomorphic on $\mathbb{C}\setminus \{0\}$. You might say "Well, it's holomorphic at $1/n$ for every $n$, so how could it fail to be holomorphic at $0$?". The answer is that we might only be able to guarantee differentiability at $1/n$ on an open ball of radius $1/n$, which then would not include $0$.  
However, there is a theorem that comes to the rescue: if, in addition, your function $f$ is bounded on a neighbourhood of $z_{0}$, then there is an analytic continuation of $f$ to all of $\mathbb{C}$. That is, there is a holomorphic function $F: \mathbb{C} \to \mathbb{C}$ which agrees with $f$ at every point in $\mathbb{C} \setminus \{z_{0}\}$.
A: See Riemann's theorem on removable singularities, whose main point is

A holomorphic function $D \setminus \{z_0\} \to \mathbb C$ has a holomorphic extension to the whole of $D$ iff it is bounded near $z_0$.

A: I think the example you should spend time thinking about is $f(z)=1/z$. The function is holomorphic on $\Bbb{C}\setminus\{0\}$, as others have already pointed out. So, your question seems to be that if it is holomorphic at all points arbitrarily close to $0$, hence in neighborhoods of all these points, then why do none of these neighborhoods contain $0$?
Well, I guess the way you could think of that is just the Hausdorffness of $\Bbb{C}$. Given any point $z\neq0$, I can find a neighborhood of $U$ of $z$ which does not contain $0$. This will be our neighborhood of $z$ on which $f$ is holomorphic. 
(Note in fact we didn't even need the complete Hausdorff property, just the fact that $\Bbb{C}$ is $T_1$)
If you have a specific follow up question, let me know and I'll try to answer it. Complex analysis feels very unintuitive to me in a lot of ways, so I understand how you feel
