The closure of the set of all permutations with finite support in $Sym( \mathbb{N})$

Question

https://www.math.tu-dresden.de/~bodirsky/lehre/alg-strukturen-ss-2016/script.pdf

Consider $Sym( \mathbb{N})$ as a topological group.

Let $P$ be the set of permutations $f$ of $\mathbb{N}$ that have finite support, i.e. $|\{i \in \mathbb{N} | f(i) \neq i\} |$ is finite.

In this lecture notes, Example 1.4.2 says that the closure of $P$ is the whole symmetric group.

Here a set $S$ is closed if it contains all $f \in Sym(\mathbb{N})$ such that for all finite $F \subset \mathbb{N}$ there exists $g \in S$ such that $f(x)=g(x) \forall x \in F$.

So my question is how do you show that $P$ is not closed and that the closure of $P$ is the whole group.

Answer 1

Consider a non-empty basic open set $S(\bar a, \bar b)$ of $\operatorname{Sym}(\mathbb{N})$, where $\bar a = (a_{1}, \dots, a_{k})$ and $\bar b = (b_{1}, \dots, b_{k})$ for some $k$. Consider $$ \Delta = \{ a_{1}, \dots, a_{k}, b_{1}, \dots, b_{k} \} $$ and the map $\sigma$ which is defined as

• first of all as $\sigma(a_{i}) = b_{i}$ for $i = 1, \dots, k$,
• then extended to a bijection of $\Delta$ (one can do this because as $S(\bar a, \bar b)$ is non-empty, we have $b_{i} \ne b_{j}$ for $a_{i} \ne a_{j}$), and
• finally extended to the identity on $\mathbb{N} \setminus \Delta$.

Then $\sigma \in S(\bar a, \bar b)$ has support $\subseteq \Delta$.