How can I find the set of limit points in this metric space? 
We have $d: {\rm I\!R}^{2}$x$ {\rm I\!R}^{2}$ -> ${\rm I\!R}$ defined by:
$d((x_1, y_1), (x_2, y_2)) := \begin{cases}|y_1-y_2|, \text{if} \ x_1 = x_2\\|y_1| + |x_1-x_2|+|y_2|, \text{if}  \  x_1 \neq x_2\end{cases}$
$M :=$ {$(t, 2) | 0 < t < 2$}

How can I find the set of all limit points of $M$? I dont't understand the relationship between $M$ and neighbourhoods.
 A: The limit points (also called the accumulation points) of $M$ are defined by: $x$ is an accumulation point of $M$ if $x\in \overline{M\setminus\{x\}}$.  In a way it's saying that $x$ is essential to the set.  In the language of neighbourhoods, $x$ is an accumulation point if any neighbourhood $U(x)$ contains infinitely many points of $M$.  Note that $x$ does not have to belong to $M$ to be an accumulation point of it.
As an example, consider the set $X:=B(0,1) \cup \{2\}$ as a set in $\mathbb R$ with the Euclidean metric $|.|$; both $\{0\}$ and $\{1\}$ are accumulation points of $X$ (consider any neighbourhood of $\{0\}$: it takes the form $(0-\epsilon, 0+\epsilon)$ which clearly intersects $X$ for any $\epsilon >0$), but $\{2\}$ is not – it is an isolated point. 
For your question then, we need to know what the neighbourhoods of a point look like.  Because you have a metric space these are the $\epsilon$-balls as determined by the metric $d$.
The distance from any point of $M$ to some arbitrary point $z=(z_1,z_2)$ is given by $d((t,2),(z_1,z_2))$ where $0<t<2$.  There are two cases to consider:


*

*$t=z_1$:  Then $d((t,2),(z_1,z_2)) = |z_2-2|$ and this can be made arbitrarily small by letting $z_2$ get close to $2$ (from either side).  If you sketch this, a small "tube" of width $2\epsilon$ around $M$ is formed by these balls.

*$t \not= z_1$: Then $d((t,2),(z_1,z_2)) = 2 + |z_2| + |t-z_1|$ and this is clearly greater than or equal to $2$ no matter what point $(z_1,z_2)$ is taken to be.  For $\epsilon <2$ then there are no points in the $\epsilon$-ball except those where $t=z_1$.


So, this means that all points $(t,2)$ with $0<t<2$ are limit points of $M$ and no other points are.  Thus $M$ is a closed set in this metric space.  (However, you should also note that the set of points $(t,0)$ with $0<t<2$ does have $(0,0)$ as a limit point, so not all sets of the form $(x,c)$ for some constant $c$ are closed.)
