Given $M=\{1,2,3,4\}$ find a topology on $M$ of minimum $3$ elements which makes $x=\{1,2,1,2,\dots\}$ converge. So, I've been having trouble with understanding this exercise:
It sounds like this:
Given $M=\{1,2,3,4\}$ find a topology on $M$ of minimum $3$ elements which makes $x=\{1,2,1,2,\dots\}$ converge.
I don't really understand what is $x$ supposed to be, and how to test it's convergence on the topology without a function.
 A: Let's apply the definition of convergence in a topological space to this example. In order for the sequence $s = (1,2,1,2, \ldots)$ to converge in some topology $\tau$ over $M$ there has to be a value $x \in M$ such that for any open set $O \in \tau$ with $x \in O$ there is some $n \in \mathbb N$ such that for all $m \ge n \colon s(m) \in O$.
If $\tau = \{ \emptyset, M \}$, this is true for $x = 1$ and $x = 2$: Consider the case $x = 1$. The only open set in $\tau$ containing $1$ is $M$ and $s(n)$ is in $M$ for every $n \in \mathbb N$. But $\tau$ only contains two elements and hence doesn't satisfy the requirements.
So consider $\tau = \{ \emptyset, \{1,2\}, M\}$. This is a topology over $M$ and I'll leave it to you to prove that the sequence $s = (1,2,1,2, \ldots )$ converges to both $x = 1$ and $x = 2$ in this topology. (Just apply the definition of convergence in a topological space.)
A: Recall that a sequence $\{x_n\}$ converges to some $x$ in a topological space if for every open set $U$ with $x\in U$, there is some $N$ sufficiently large that $x_n \in U$ whenever $n>N$.
If the sequence $\{1,2,1,2,\dotsc\}$ is to converge to some $x\in M$, then every open set containing $x$ must contain both 1 and 2, since both of these terms appear infinitely often in the sequence.  In particular, if we take the indiscrete topology on $M$ (i.e. the only open sets are the empty set and $M$ itself), then the sequence converges to anything we like.  However, the question demands that we have at least three sets in the topology.
Note that if either $\{1\}$ or $\{2\}$ is open, then the sequence won't converge (do you see why?).  However, we could throw in the set $\{1,2\}$.  That is, if the collection
$$ \mathcal{T} = \{ \emptyset, \{1,2\}, M\} $$
is a topology on $M$, then the sequence $\{1,2,\dotsc\}$ will converge (to both 1 and 2), and we will be done.
So, is $\mathcal{T}$ a topology?  It contains the empty set and $M$, it is closed under arbitrary unions, and it is closed under finite intersections, so yes!  it does form a topology!  Hence one possible answer to your question is the collection $\mathcal{T}$, above.  Can you come up with another example (by, perhaps, throwing in more sets)?
A: You can take the  topology $\mathcal{T}=\{M, \emptyset\,\{1,2\}\}$ 
Use the definition of the convergence of a sequence in a topological space to see why this topology is a valid one.
