TITLE: Can the topology $\tau = \{\emptyset, \{a\}, \{a,b\}\}$ be induced by some metric?
QUESTION [6 upvotes]: I'm just beginning to learn about topological spaces, and in my notes a few examples of topological spaces are given. 
One of them is $\tau = \{\emptyset, \{a\}, \{a,b\}\}$ on the set $X = \{a,b\}$. I can clearly see how this meets my definition of a topology, namely that 

$\emptyset, X \in T$;
$\bigcup_{\lambda \in \Lambda} T_{\lambda} \in \tau$ whenever $\{T_{\lambda}\}_{\lambda \in \Lambda} \subseteq \tau$;
$\bigcap_{k=1}^{n} T_{k} \in \tau$ whenever $\{T_{k}\}_{k=1}^{n} \subseteq \tau$.

My question: I was wondering if this could be a topology induced by some metric $d$ on $X$, however. Or more generally, how one could determine whether a particular topology is possibly induced by a metric. 
I have seen that the discrete topology on any $X \neq \emptyset$ given by $\tau = \mathcal{P}(X)$ is induced by the discrete metric $\mu$, for instance, which has motivated my question as to whether some of the other topologies introduced may be metric-induced as well. I attempted to think of some metric which would induce my above topology, but have thus far been unsuccessful - but also not sure how I might prove that one cannot exist.

REPLY [7 votes]: Every metric space is Hausdorff. This topology is not. Every neighbourhood of $b$ contains $a$. So $\tau$ is not induced by any metric.

REPLY [5 votes]: It is easy to prove that every topology $\tau$ induced by a metric space $(X,\delta)$ on a finite set $X$ is discrete.
How? Just prove every point is open, to do this notice that $\{x_0\}=B(x_0,d)$. Where we define $d$ as $\min\limits_{x\in X-\{x_0\}} \delta(x,x_0)$