A question about metrizability In a lecture in Topology I had earlier this week, I was told (without proof) that not every topological space $(X,O)$ is metrizable, i.e, it is impossible to find some metric $d$ such that  $O$  and $O(d)$ are topologically equivalent. I've been trying to come up with a counter example but it's a bit hard to visualize. Does anyone know a reference or have a proof of this theorem?
 A: There are certain topological properties that all metric spaces share:


*

*They are all Hausdorff (distinct points can be separated by disjoint open sets) and perfectly normal (normal means disjoint closed sets can be separated by disjoint open sets while the "perfectly" modifier means that additionally all closed sets are Gδ).

*The are all first-countable (each point has a basis of neighbourhoods consisting of only countably many sets).

*They are all paracompact (for any open cover $\mathcal{U}$ you can find another open cover $\mathcal{V}$ with the property that every set in $\mathcal{V}$ is a subset of some set in $\mathcal{U}$, and every point has a neighbourhood which meets only finitely many members of $\mathcal{V}$).

*If a metric space satisfies one of the following properties, it satisfies all of them:


*

*Separability (there is a countable dense set).

*Second-countability (there is a countable base).

*Lindelöfness (every open cover has a countable subcover).

*ccc-ness (every collection of pairwise disjoint open sets is countable).


(There is an analogous property for any infinite cardinal $\kappa$.)

*If a metric space satisfies one of the following properties, it satisfies all of them:


*

*Compactness (every open cover has a finite subcover). 

*Countable compactness (every countable open cover has a finite subcover).

*Sequential compactness (every sequence has a convergent subsequence).



The above is not a complete list, but any space disobeying at least one of the above is not a metric space.
For example, the long line has the following properties which show it is not metrizable:


*

*it is a normal Hausdorff space which is not perfectly normal.

*it is sequentially compact and countably compact but not compact (or even Lindelöf).

*it is not paracompact.

A: Here is a good example I keep in mind.  There are minor things to check, but I like working with the familiar object $\mathbb{R}$.
Consider $\mathbb{R}$ with two different topologies: the "normal" topology on $\mathbb{R}$ (basis is given by open intervals $(a,b)$).  $\mathbb{R}$ with this topology is metrizable of course ($d(x,y) = |x-y|$).
Now, let's endow $\mathbb{R}$ with a different topology (called the lower limit topology) where the basis for open sets in $\mathbb{R}$ are given by intervals of the form $[a,b)$ (you should check that this is actually a basis).  Let's denote $\mathbb{R}$ with this topology as $\mathbb{R}_\ell$.
The two topologies $\mathbb{R}$ (standard one) and $\mathbb{R}_\ell$ looks similar, but you should convince yourself that $\mathbb{R}_\ell$ many many more open sets.  In fact, here is the crucial observation you should make: $\mathbb{R}$ with standard topology admits a countable basis (take all open intervals of rational radius centered at rational points), but $\mathbb{R}_\ell$ does not (you should prove, or at least convince yourself of this).
Now, a metric space with countable dense subset always admits a countable basis (like $\mathbb{R}$ with standard topology as we just discussed).  However, $\mathbb{R}_\ell$ has a countable dense subset but does not admit a countable basis, hence it must not be metrizable.
References: Munkres' Topology, my past topology class notes.
A: Every compact metrizable space is second-countable.  But the ordered square $[0,1]^2$ with the order topology induced  by the dictionary order is compact since it is a closed interval in its order.  However, it is not second countable, and therefore is not metrizable.
A: Let $X = \{a,b,c\}$ and let $\tau = \{\emptyset,X\}$ be a topology on $X$ (this is called the trivial topology on $X$). Then suppose (for a contradiction) that $(X,\tau)$ is metrizable, and let $d:X \to [0,\infty)$ be a metric on $X$ such that its metric topology coincides with $\tau$.
Then $U = \{x \in X : d(x,a) < d(a,b)\}$ is an open set. Note $b \not\in U$,
but $a \in U$, so we either have $U = \{a,c\}$ or $U = \{a\}$, neither of which are in $\tau$, and hence are not open, a contradiction.
