# Help with metrics, Box topology and non-metrisable Hausdorff spaces

I'm trying to come up with a simple example of a second countable Hausdorff space that is not metrisable. The most promising I've come up with so far is the Box topology:

The following is a basis for the Box topology on $\mathbb R^{\mathbb N}$: $\{ \prod_{n \in \mathbb N} O_n \mid O_n \text{ open in } \mathbb R \}$.

Unfortunately, I have been unable to either put a suitable metric on it or show that it's not metrisable. The metrics that have come to mind so far are:

(i) $d_n (x,y) = (\sum_k |x_k - y_k|^n)^{\frac{1}{n}}$, inducing the product topology and

(ii) the discrete metric inducing the discrete topology

Question 1: What other metrics (equivalent to the above or not) exist that I have not thought of?

Question 2: If it turned out that this is not metrisable: how does one show that a space is not metrisable? (in general, or if there is not general recipe, in this case)

Question 3: Can you come up with a simpler example than the Box topology?

Many thanks for the help.

-
What about $\sup_n |x_n - y_n|$? –  Ilya Jan 15 '13 at 18:56
For examples, try using this website. –  David Mitra Jan 15 '13 at 18:57
@Ilya, you need to bound that metric. –  JSchlather Jan 15 '13 at 18:59
@MattN.: oh, as Jacob advised instead of $|\cdot|$ please take $\frac{|\cdot|}{1+|\cdot|}$. And don't forget to check whether this metric indeed induces the very same topology –  Ilya Jan 15 '13 at 19:01
The box topology is not first countable, so it is not metrizable. –  Martin Jan 15 '13 at 19:02

A second countable Hausdorff space is metrizable iff it is regular, so you needn’t even worry about metrics: you just need a non-regular Hausdorff space with a countable base. One easy example is to let $X=\Bbb Q^2$ with the following topology. Points $\langle p,q\rangle$ with $q\ne 0$ are isolated. A basic open nbhd of $\langle p,0\rangle$ is any set of the form

$$\big\{\langle p,0\rangle\big\}\cup\left((p-\epsilon,p+\epsilon)\times\Big((-\epsilon,\epsilon)\setminus\{0\}\Big)\right)\;.$$

In other words, points on the $x$-axis have nbhds consisting of themselves and the intersection of an ordinary Euclidean nbhd with the complement of the $x$-axis.

Added: Box products are not the way to go. Suppose that there are distinct points $x,y\in X$ such that $D=\{x,y\}$ is a discrete subspace of $X$. Then the box product $\square X^\omega$ of $\omega$ copies of $X$ contains $\square D^\omega$ as a discrete subspace of cardinality $2^\omega$ and therefore cannot be second countable.

Added2: Let $S=\{0\}\cup\left\{2^{-n}:n\in\Bbb N\right\}$ as a subspace of $\Bbb R$. Let $Y=\Bbb N\times S$, let $p$ be any point not in $Y$, and let $X=\{p\}\cup Y$. $Y$ is an open subspace of $X$. For $n\in\Bbb N$ let

$$B_n(p)=\{p\}\cup\big\{\langle k,s\rangle\in Y:k\ge n\text{ and }s\ne 0\big\}\;,$$

and take $\{B_n(p):n\in\Bbb N\}$ as a local base at $p$. $X$ is countable and first countable, so $X$ is second countable, and $X$ is clearly Hausdorff. However, $\Bbb N\times\{0\}$ is a closed subset of $X$ that cannot be separated from $p$ by disjoint open sets, so $X$ is not $T_3$.

-
Thank you! You're very helpful. I will accept it tomorrow after re-reading it. Now I'm too tired. I hope you don't mind. (I already upvoted) –  Rudy the Reindeer Jan 15 '13 at 19:35
@Matt: No problem. I might even think of another good example. :-) –  Brian M. Scott Jan 15 '13 at 19:38

Another example is the $R_K$ topology. If $K$ denotes the set of all numbers of the form $\frac{1}{n}$ for $n \in \mathbb{Z}_+$, it is the topology generated by all open intervals $(a.b)$ together with those of form $(a,b) - K$. So it's in spirit related to Brian Scott's example (actually, 2 of them), but - for all I know - one of the 2 or 3 standard easy examples to look at for unusual behavior (eg, as Munkres introduces point set topology, it keeps being discussed (text and exercises); it is also in Counterexamples in Topology)).

As in BMS's examples, the $R_K$ topology is not regular but 2nd countable, and so cannot be metrizable. To see this, note:

(a) Not regular:

The set $K$ is closed, and it cannot be separated from $0$: assuming two open sets $U$ (containing $0$) and $V$ (containing $K$) separate them, then there is an open neighborhood around $0$ in $U$ which must be of form $(a,b) - K$ (or it intersects $K$); choose $n$ large enough that $\frac{1}{n} \in (a,b)$, and a neighborhood around $\frac{1}{n}$ in $V$ which must be of form $(c,d)$, chosen such that $c > 0$, and $(c,d) \subset (a,b)$. Then it is easy to see (eg, draw the picture of the above) that there is a point in $U \cap V$ ("between $c$ and $\frac{1}{n}$").

(b) 2nd countable:

This is rather obvious (for the same reason that the standard topology is 2nd countable)

-

You might want to check out Example 2 in Topology (2nd ed.) by Munkres on p. 133. It shows that an uncountable product of $\mathbb{R}$ cannot be metrizable since it fails the sequence lemma given on p. 130 of this textbook.

-