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The Theorem: Let $G$ be a topological group. Then $G$ is metrizable iff $G$ is Hausdorff and the identity $1$ has a countable neighborhood basis. Moreover, if $G$ is metrizable, $G$ admits a compatible metric $d$ which is left-invariant: $d(xy, xz)=d(y, z)$.

Source: Classical Descriptive Set Theory by Alexander S. Kechris.

My Question: The theorem was given in the book without a proof. What exactly is a "neighborhood basis"? It is clear to me that metrizable implies Hausdorff, but that is about the furthest I could go. Any help would be greatly appreciated.

Asaf Karagila
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Dick Grayson
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2 Answers2

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See, Understanding the concept of neighborhood basis

Neighbourhood basis $N$ about a point $x$ means a collection of subsets of $G$ which is equal to $N$ such that if $V$ is any open set containing $x$ there exists an open set $U \in N$ such that $U \subseteq V$.

Now if we are able to choose $N$ with countable number of subsets then we say the neighbourhood basis about $x$ is countable.

Some comments: $d(xy,xz) \implies x \ is \ an \ isometry$. i.e., group operation is an is0metry.

Some examples:

An example: is $G = U(n)$ the space of $n \times n $ unitray matrix where $d(U_1,U_2) = ||U_1 - U_2||$ works.

In case of complex numbers the group operation works like $z_1 * z_2 = z_1+z_2$. Hence $d(z*z_1,z*z_2) = |z_1-z_2| = d(z_1,z_2)$

In particular if $G$ is a Vector space with a norm i,e., a Banach space, then $d(x,y) = ||x-y||$ with $x*y = x+y$ will be the theorems output metric.

Countable Basis about $x$ for all above examples: $$=\{B_r(x): r \ is \ a \ rational \ number \}.$$ $$B_r(x) = \{y: ||x-y|| < r\}.$$

This is a countable basis since any open set containing $x$ must contain $B_r(x)$ for some $r$.

In general, By assuming countable basis at unity of the group the authors assume countable basis for all points as if $V$ is a neighbourhood of $1$ then $xV$ is a neighbourhood around $x$ and vice-versa by multiplying by $x^{-1}$. So they are saying the space is first countable.

See: https://en.wikipedia.org/wiki/First-countable_space

So the authors say $G$ is metriziable with group operation as isometry iff $G$ is Hausdorff and is first countable.

Balaji sb
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A neighbourhood basis of a point $x\in G$ is a collection $\mathcal{C}$ of open neighbourhoods of $x$ such that for every open set $U$ containing $x$, there is some $V\in \mathcal{C}$ such that $V \subseteq U$. In a metric space, every point $x$ admits a countable neighbourhood basis, namely $\{B(x, r):r\in \mathbb{Q}\}$.

I'll outline the proof of Birkhoff Kakutani. Let $G$ be a topological group such that the identity 1 admits a countable neighbourhood basis $\mathcal{C}=\{U_n : n\in \mathbb{N}\}$. Argue that we can choose this neighbourhood basis such that $U_{n+1} \cdot U_{n+1} \subseteq U_n$ for all $n$. For a point $x\in G$, we define its distance from 1 as follows: \begin{align*} d(x, 1) = \inf\{2^{-n_1}+...+2^{-n_k} : x\in U_{n_1} \cdot \cdots \cdot U_{n_k}\}. \end{align*} (take $U_0=G$ to ensure this is well-defined). Try to picture why this is a good notion of distance. Why is $d(x, 1)$ larger when $x$ is "far" from 1, and smaller when $x$ is "close" to 1?

Now that we know how distance works around the identity, we can use the group operation to find the distance between any two points: \begin{align*} d(x, y) = d(y^{-1}\cdot x, 1). \end{align*} Why is this a reasonable definition? Hint: picture the case where $G$ is a topological vector space.

Now you still have to show that $d$ is indeed a metric (a left-invariant metric, moreover), and that this metric induces the right topology. I leave that to you (this is non-trivial).

JMM
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