Show that the topological space ( X, $\tau$ ) is not metrizable For the topological space ( X, $\tau$ ), with X = {0, 1} and $\tau$ = { $\emptyset$ , {0}, {0,1} } , prove that ( X, $\tau$ ) is not metrizable. 
I know intuitively it can't be but don't know how to prove it. The only idea I have is that {0, 1} cannot be open as X = {0, 1} is discrete but I'm not even sure if this is right. 
 A: You can use the fact that any metrizable space is Hausdorff. This follows the definition of a metric, in particular the non-negative axiom ($d$ is a metric implies that $d(x,y)\geq0$ and $d(x,y)=0$ if and only if $x=y$).
Let me know if you'd like me to flesh out the details.

Edit: A topological space $(X,\tau)$ is Hausdorff if and only if for any two points $x,y$ in $X$ you can find two sets $U$, $V$ in $\tau$ such that $x\in U$, $y\in V$ and $U\cap V=\emptyset$ (that is, if you can "house" any two points in disjoint open sets).
This is clearly not the case in your example, since you cannot find such sets for $x=0$ and $y=1$ (try!).
So we know that $(X,\tau)$ is not Hausdorff. If we can show that any metrizable space must be Hausdorff then it follows that $(X,\tau)$ is not metrizable.
Suppose that $(X,\tau)$ is metrizable, pick a metric $d$ and any two points $x,y\in X$. Let $\varepsilon=\frac{1}{2}d(x,y)>0$ (by the metric axioms).
Then $U:=B_\varepsilon(x)$ and $V:=B_\varepsilon(y)$--where $B_\varepsilon(x)$ denotes the open ball of radius $\varepsilon$ centred on $x$--are open, disjoint and $x\in U$, $y\in V$. Since, $x,y$ where arbitrary we have that $(X,\tau)$ is Hausdorff.
A: If $d$ would be such a metric, $B(1,d(0,1))$ would be open and equal to $\{1\}$. Contradiction.
More generally, a finite topological space is metrizable iff it is discrete.
($\Rightarrow$) If $X=\{x_1,\ldots,x_n\}$ and $d$ is a metric on $X$, then $\varepsilon=\min\{d(x_i,x_j):i\neq j\}>0$ and $B(x_i,\varepsilon)=\{x_i\}$ is open for every $i$, hence every subset $U$ of $X$ is open : 
$$U=\bigcup_{x\in U} \{x\}$$
($\Leftarrow$) The discrete topology arises from the metric  $$d:(x,y)\mapsto\left\{\begin{array}{rl}1 & \text{if}\ x\neq y \\ 0 & \text{if}\ x=y \end{array}\right.$$ 
on any set $X$.
A: An easy proof uses the following: A finite topological space is $T1$ (and anything stronger, like metrizable) iff its topology is the discrete topology.
Assuming this, we see that $\tau = \{ X , \emptyset, \{0\}\}$ is not metrizable because it's not the discrete topology (it's missing $\{1\}$ as an open set).
