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\textbf{MATH 535, Topology, Homework 2}
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\underline{\textbf{1. (2+1 pts)}} Let $(X_i,\tau_i)$ be separable topological spaces for all $i \in \mathbb{N}$.\\

a) Show that $\displaystyle \prod_{i \in \mathbb{N}} X_i$ together with the product topology is separable.\\

b) Show that $\displaystyle \prod_{i \in \mathbb{N}} \{0,1\}$ together with the box topology where each component has the discrete topology is not separable.\\


\underline{\textbf{2. (2+1+4 pts)}}\\

Recall that the Baire space $\mathbb{N}^{\mathbb{N}}$ is the set of functions from $\mathbb{N}$ to $\mathbb{N}$ whose topology is induced by the metric
\[ d(f,g)=\displaystyle \frac{1}{\min\{i \in \mathbb{N}: f(i) \neq g(i)\}+1}\]
It follows from a theorem we proved in class that the Baire space is a complete metric space. We can consider the infinite symmetric group $Sym(\mathbb{N}) \subseteq \mathbb{N}^{\mathbb{N}}$ as a subspace of $\mathbb{N}^{\mathbb{N}}$. In this case, the subspace topology of $Sym(\mathbb{N})$ is induced by the restriction of the metric $d$ to $Sym(\mathbb{N}) \times Sym(\mathbb{N})$.\\

a) Let $f_n=(0,1,\dots,n)$ for each $n \in \mathbb{N}$. Here we use the standard cycle notation, that is, $f_n: \mathbb{N} \rightarrow \mathbb{N}$ is the element of $Sym(\mathbb{N})$ which sends $0$ to $1$, $1$ to $2$,... and $n$ to $0$.\\

Show that the sequence $(f_n)_{n \in \mathbb{N}}$ is a Cauchy sequence which is not convergent in the metric space $(Sym(\mathbb{N}),d)$.\\

b) Is the subspace $Sym(\mathbb{N})$ completely metrizable?\\



For the remainder of this question, let us first recall the definition of a \textbf{topological group}. A topological group is a group $G$ together with a topology, with respect to which the multiplication operation $(g,h) \mapsto g \cdot h$ from the product space $G \times G$ to $G$ and the inversion operation $g \mapsto g^{-1}$ from $G$ to $G$ are continuous maps. For example, the group $(\mathbb{R},+)$ is a topological group since the addition map $(x,y) \mapsto x+y$ from $\mathbb{R}\times \mathbb{R}$ to $\mathbb{R}$ is a continuous map and the map $x \mapsto -x$ from $\mathbb{R}$ to $\mathbb{R}$ is also a continuous map. Recall that $Sym(\mathbb{N})$ together with composition of functions as the multiplication forms a group.\\

c) Prove that $Sym(\mathbb{N})$  is a topological group.\\

(\textbf{Hint.} First try to argue that composition of bijections is continuous as a map from $Sym(\mathbb{N}) \times Sym(\mathbb{N})$ to $\mathbb{N}^{\mathbb{N}}$. Similarly, argue that inversion is continuous as a map from $Sym(\mathbb{N})$ to $\mathbb{N}^{\mathbb{N}}$. Then argue that these maps will stay continuous if we shrink the codomains to $Sym(\mathbb{N})$. In order to do the former, ask yourself this: When is a function from a topological space to a product space continuous? Try to remember the theorem we proved in class regarding this. It may give you a faster solution than trying to use the definition of continuity or the equivalent $\epsilon$-$\delta$ condition on the whole product space.)
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