Countability Axioms and Topological Groups (4.30.18)

James Munkres Topology, chapter 4.30, exercise 18:

Show in a first-countable topological group $G$ that second countability, separability, and the Lindelof condition are equivalent.

Proof: Let $\{B_n\}$ be a countable basis about $e$. We may presume $B_n⊇B_{n+1}$ for all $n$, and if necessary by setting $B_n'=B_n∩B_n^{-1}$ we may presume $B_n^{-1}=B_n$.

Suppose $G$ has a countable dense subset $D$, and that $U$ is a hood about $g$. Then since $g×e∈m^{-1}(U)$ and $\{gB_n\}$ is a basis about $g$, there exists $n$ such that $gB_n×B_n⊆m^{-1}(U)$. Choose $d∈D∩gB_n$ and we see $d=gb$ for some $b∈B_n$, hence $g=db^{-1}∈dB_n$, so $dB_n$ is a hood about $g$. Furthermore, given $c∈B_n$ we see $dc=gbc∈m(gB_n×B_n)⊆U$, so that $dB_n⊆U$. Therefore $\{dB_n~|~d∈D,n∈ℕ\}$ is a countable basis for $G$.

Suppose $G$ is Lindelof. Then for each $n∈ℕ$, there is a countable subset $D_n⊆G$ such that $∪_{d∈D_n}dB_n=G$. We show $D=∪D_n$ is dense in $G$; let $U⊆G$ be a hood about $g$. Then $gB_n⊆U$ for some $n$, and we also see $g∈∪_{d∈D_n}dB_n$ so write $g=db$ for some $d∈D_n⊆D$ and $b∈B_n$. But now $d=gb^{-1}∈gB_n⊆U$ so $d∈U$ and $D$ is a countable dense subset of $G$.$~\square$

Separability of Large Euclidean Product Spaces (3.30.16)

James Munkres Topology, chapter 3.30, exercise 16:

(a) Show that the product space $ℝ^I$ contains a countable dense subset.
(b) Show that if $|J| > |\mathcal{P}(ℕ)|$, then $ℝ^J$ does not contain a countable dense subset.

Proof: (a) For every oddly finite sequence $q_1,...,q_{2n+1}$ of rationals such that $0 ≤ q_2 < q_4 < ... < q_{2n} ≤ 1$, let there be an associated function $I → ℝ$ that takes value $q_{2k-1}$ on $(q_{2k-2},q_{2k})$ for each $k = 1, ..., n$ (where $q_0=0$), takes value $q_{2n+1}$ on $(q_{2n},1]$, and is zero on each of $q_2,...,q_2n$. We see the collection of such functions is countable, and we claim it forms a countable dense subset of $ℝ^I$: For a given basis element $∏_{i∈I}U_i$ of open sets $U_i⊆ℝ$ such that $U_i \not = ℝ$ for only finitely many $i∈I$, let $i_1 < ... < i_n$ be exactly all such that $U_{i_j} \not = ℝ$. We may choose rationals $q_2,...,q_{2(n-1)}$ such that $i_1 < q_2 < i_2 < q_4 < ... < q_{2(n-1)} < i_n$ (unless $n=1$ where one chooses $i_1 < q_2$ and say $q_3=0$) and for each $q_1,q_3,...,q_{2n-1}$ choose rationals such that $q_{2k-1}∈U_{i_k}$ for each $k=1,...,n$. Then the associated function on $q_1,...,q_{2n-1}$ takes a value within $U_i$ on each $i$, and hence is contained in the basis element.

(b) Suppose $D$ is a countable subset of $ℝ^J$. Fix some nonempty interval $(a,b)$, and choose some disjoint nonempty interval $(c,d)$. Then since the function $f : J → \mathcal{D}$ given by $f(α) = D ∩ π_α^{-1}(a,b)$ cannot be injective, we find some distinct $α,β∈ℝ$ such that for each $d∈D$ we have $d(α)∈(a,b)$ iff $d(β)∈(a,b)$. But now the basis element $∏U_j$ where $U_α=(a,b)$ and $U_β=(c,d)$ and $U_j=ℝ$ otherwise cannot contain a point of $D$.$~\square$