Are ultrafilters unique? I'm trying to get a feel for what an ultrafilter is by looking at some finite examples. Firstly, I took a look at this question: Example of a filter on a set, and it stoked the question: can sets such as the one in this question have several ultrafilters?
As an example of what I had been thinking: the answer to the aforementioned question states that the given ultrafilter is "centered at $3$", and so it follows that we should be able to center the filter at, for example, $4$, as well. In which case we would have that $f(S)=\lbrace\lbrace4\rbrace,\lbrace3,4\rbrace,\lbrace4,5\rbrace,\lbrace3,4,5\rbrace\rbrace$. Is this correct, or is there a flaw in my reasoning?
 A: It's useful to have the following characterisations of ultrafilters on a set $X$:
For a  non-trivial filter $\mathcal{F}$ on $X$, the following statements are equivalent: 


*

*$\mathcal{F}$ is an ultrafilter (i.e. maximal w.r.t. inclusion) 

*For all finitely many $A_1,\ldots A_n \subseteq X$ we have that $\cup_{i=1}^n A_i \in \mathcal{F}$ iff there is some $1 \le i \le n$ such that $A_i \in \mathcal{F}$.

*For all subsets $A$ of $X$ either $A \in \mathcal{F}$ or $X\setminus A \in \mathcal{F}$.


Equivalence 2. shows that if $\mathcal{F}$ contains a finite subset $F=\{x_1,\ldots,x_m\}$ it contains one of the sets $\{x_i\}$, and then $\mathcal{F} =  \{A: x_i \in A\}$ for that $x_i$, i.e. $\mathcal{F}$ is a principal filter. IN particular on a finite set $X$ we only have the principal ultrafilters.
So a non-principal filter cannot contain any finite subset $F$, so by 3. $\mathcal{F}$ contains all co-finite subsets, i.e. $\mathcal{F}$ is an enlargement of the co-finite filter. 
So ultrafilters are of two types: the principal ones (of which there are as many as points of $X$) and the free ones, that we can "get" by invoking choice and somehow extending the cofinite filter to an ultrafilter. There are $2^{2^{|X|}}$ of those for an infinite set $X$ (so many many more), assuming choice. But these are non-constructive and we can try to visualise them as some way of choosing for each infinite subset $A$ of $X$ whether to put $A$ in $\mathcal{F}$ or its complement, but in a "filter-consistent" way (if we put in a set, we also have to put in all larger sets, and if we put in two, we also have to add the intersection) and no finite set is allowed in $\mathcal{F}$ if we want a free ultrafilter, so their complements are surely in. It's not much but maybe it helps.
