# Can the topology $\tau = \{\emptyset, \{a\}, \{a,b\}\}$ be induced by some metric?

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.

• Well, have you tried listing all the possible metrics on $X$, and see what topologies they give? This isn't so hard, since $X$ has only two points... – Eric Wofsey Oct 18 '16 at 23:52
• Hint: consider two cases, $d(a,b)=0$ or $d(a,b)\gt0.$ – bof Oct 18 '16 at 23:57
• @bof In a metric space $d(a,b)=0$ iff $a=b$. – Cheerful Parsnip Oct 19 '16 at 0:02
• @bof and Eric - thanks for the hints, it is making much more sense to me now. – Eric Hansen Oct 19 '16 at 0:02
• @GrumpyParsnip I know. I was thinking of showing that the topology is not induced by a pseudometric. – bof Oct 19 '16 at 0:07

Every metric space is Hausdorff. This topology is not. Every neighbourhood of $b$ contains $a$. So $\tau$ is not induced by any metric.
• Better: Every pseudometric space is regular. This topology is not. The neighbourhood $\{a\}$ of the point $a$ contains no closed neighbourhood of $a.$ So $\tau$ is not pseudometrizable. (Or uniformizable.) – bof Oct 19 '16 at 0:01
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 $\min\limits_{x\in X-\{x_0\}} \delta(x,x_0)$