Let $N$ be a condensing submodule. We have a homomorphism of the modules $M_i$ into $M/N$, and the union of their images is all of $M/N$. Conversely, if there is such a homomorphism of the $M_i$ into a module $K$ with the union property, then we have a surjective homomorphism $φ : M → K$ given by evaluation of the sum of the nonzero components in $K$, giving rise to a condensing submodule as the kernel. The equivalence regarding halt-condensing submodules is modified by requiring the mapped modules to inject and be disjoint at nonzero values.
Let $R$ be a UFD with group of units $V$. Choose a list of primes $p_i$ of cardinality $|S|=c$ to represent the associativity classes of prime elements. Letting $A=\text{Aut}(V)$ (group automorphisms), $S_c$ be the symmetric group on $c$ objects, and $V^*$ be the group of maps $S→V$ where the operation is multiplication $φ_1φ_2(s)=φ_1(s)φ_2(s)$, we claim $A × S_c × V^*$ contains (a copy of) $\text{Aut}(R)$ (ring automorphisms) where the former is under a custom operation elucidated below.
Let $φ∈\text{Aut}(R)$. Since automorphisms map units to units, we must have $φ|_V∈\text{Aut}(V)$. Now automorphisms also map primes to primes, and more generally associativity classes of primes to other associativity classes, so $φ$'s action on $p_i$ might be completely represented by an element $σ$ from $S_c$ together with a map $ψ:S→V$ given by $ψ(s)=φ(p_s)/p_{σ(s)}$ (i.e. so we may construct $φ$ from $ψ$ and $σ$ by $φ(p_s)=ψ(s)p_{σ(s)}$). Now, $φ$ is uniquely determined by this action on the units and primes, so we have an injection $\text{Aut}(R)$ into $A × S_c × V^*$. In order to claim an algebraic embedding the structure imposed on the latter is not simply componentwise multiplication but is rather modified to imitate composition of automorphisms of $R$. The operation is defined as such:
Let $x_1,x_2=(φ_1,σ_1,ψ_1),(φ_2,σ_2,ψ_2)∈A × S_c × V^*$. Then $x_1x_2=(φ_1∘φ_2,σ_1∘σ_2,(φ_1∘ψ_2)·(ψ_1∘σ_2))$, where multiplication and composition of maps are here distinct.
Let $A,B$ be groups. Consider $\text{Aut}_A(A \times B)$, the set of automorphisms of $A \times B$ mapping $A \times 1$ to $A \times 1$. This is seen to be a subgroup of $\text{Aut}(A \times B)$ which we shall now classify. Let $\Phi \in \text{Aut}_A(A \times B)$. By definition $\Phi$ restricts to an automorphism $φ$ on $A \times 1$, and by simultaneously defining $(\psi(b), \sigma(b))=\Phi(1,b)$ we obtain homomorphisms $\psi : B \rightarrow A$ and $\sigma : B \rightarrow B$. Further, we see $\sigma$ is injective because if $\sigma(b)=1$ then $\Phi(1,b) \in A \times 1$ and the automorphism $φ$ leads to an element $\Phi(a,1)=\Phi(1,b)$ implying $a,b=1$. Further, $σ$ is surjective seeing as $Φ$ is surjective. Assume $\text{img }ψ \not ∈ Z(A)$; then $aψ(b) \neq ψ(b)a$ for some $a∈A$, $b∈B$. Then we have $$Φ((1,b)(φ^{-1}(a),1))=Φ(1,b)Φ(φ^{-1}(a),1)=$$$$(ψ(b),σ(b))(a,1) \neq (a,1)(ψ(b),σ(b)) = Φ((φ^{-1}(a),1)(1,b))$$a contradiction. Since $φ,ψ,σ$ uniquely determine $Φ$, we have an injection of sets $\text{Aut}_A(A × B) → \text{Aut}(A) × \text{Hom}(B,Z(A)) × \text{Aut}(B)$.
Conversely, let $φ∈\text{Aut}(A)$, $ψ∈\text{Hom}(B,Z(A))$, and $σ∈\text{Aut}(B)$. Define $Φ: A×B → A×B$ by $Φ(a,b)=(φ(a)ψ(b),σ(b))$. We observe $Φ((a_1,b_1)(a_2,b_2))=Φ(a_1,b_1)Φ(a_2,b_2)$, so it is homomorphic. As well, it is injective as $Φ(a,b)=(1,1)$ implies $σ(b)=1$ so $b=ψ(b)=1$ and consequently $a=1$. Finally, we observe surjectivity $(a,b)=Φ(φ^{-1}(aψ(b)^{-1}),σ^{-1}(b))$, so $Φ$ is an automorphism. Thus we have a bijection of sets $\text{Aut}_A(A × B) → \text{Aut}(A) × \text{Hom}(B,Z(A)) × \text{Aut}(B)$.
Suppose $Φ_1$ associates to $(φ_1,ψ_1,σ_1)$ and $Φ_2$ associates to $(φ_2,ψ_2,σ_2)$. Then we observe $Φ_1Φ_2$ associates to $(φ_1φ_2,φ_1ψ_2+ψ_1σ_2,σ_1σ_2)$. Thus, defining such a binary operation on $\text{Aut}(A) × \text{Hom}(B,Z(A)) × \text{Aut}(B)$ induces a group isomorphism between the two.
Footnotes: This machinery works especially well when $Z(B)$ is manageable, and in particular when $Z(B)=1$ and $A$ is abelian we have $\text{Aut}_A(A × B) = \text{Aut}(A × B)$, as then $Z(A × B) = A × 1$ and so $A × 1$ is characteristic. Otherwise, one may possibly discover $A × 1~\text{char}~A × Z(B)~\text{char}~A × B$ and thus $A × 1~\text{char}~A × B$ to the same effect.
For groups $A≤B≤G$ let $\text{Aut}^G(A)$ be the subgroup of automorphisms of $A$ extending to automorphisms of $G$. Let $\text{Aut}_A^{G}(B)$ be the subgroup of automorphisms of $B$ mapping $A$ to $A$ and extending to automorphisms of $G$.
Let $H ≤ N \text{ char } G$. Then we have a bijection $ψ$ of coset representatives $[\text{Aut}(G)~:~\text{Aut}_H(G)]=[\text{Aut}^G(N)~:~\text{Aut}_H^G(N)]$ given by restriction $ψ(σ)=σ|_N$. This is injective as $σ_1|_N\text{Aut}_H^G(N)=σ_2|_N\text{Aut}_H^G(N)$ implies $σ_2^{-1}|_Nσ_1|_N∈\text{Aut}_H^G(N)$ so $σ_2^{-1}σ_1∈\text{Aut}_H(G)$, and surjective seeing as the representative $σ|_N$ by definition has an extension $σ$ and letting $σ'$ represent $σ$ we see $σ'|_N\text{Aut}_H^G(N)=σ|_N\text{Aut}_H^G(N)$.
Let $(X,d)$ be a metric space, and let $\mathcal{C}$ be the set of all nonempty compact subspaces of $X$. Let $A∈\mathcal{C}$. Then given $x∈X$, define $d(x,A)=d(A,x)=\min_{a∈A}d(x,a)$. Note that since $A$ is compact this is well defined. Note also that fixing $A$ this function is continuous $X→ℝ$. Hence given $A,B∈\mathcal{C}$ $$D(A,B)=\max \{\max_{a∈A} d(a,B), \max_{b∈B} d(b,A)\}$$ is well defined. It is clear $D(A,A)=0$, and $D(A,B)=0$ quickly implies $A⊆B$ and $B⊆A$. As well, $D(A,B)=D(B,A)$. Now, we show it satisfies the triangle inequality.
Let $A,B,C∈\mathcal{C}$ and assume $D(A,C) > D(A,B) + D(B,C)$. We may assume $\max_{a∈A} d(a,C) ≥ \max_{c∈C} d(c,A)$, for otherwise interchanging $A,C$ would result in such a situation.
Let $a∈A$ and $β∈B$ be such that $d(a,β)=d(a,B) ≥ d(x,B)$ for all $x∈A$, let $b∈B$ and $γ∈C$ be such that $d(b,γ)=d(b,C)≥d(x,C)$ for all $x∈B$, and let $a'∈A$ and $γ'∈C$ be such that $d(a',γ')=d(a',C)≥d(x,C)$ for all $x∈A$. Since $d(a',B)≤d(a,B)$, let $x∈B$ be such that $d(a',x)≤d(a,β)$. Since $d(x,C)≤d(b,C)$, let $y∈C$ be such that $d(x,y)≤d(b,γ)$. Now we exhibit $$D(A,C) = d(a',γ') ≤ d(a',y) ≤ d(a',x)+d(x,y)$$$$≤ d(a,β)+d(b,γ) ≤ D(A,B)+D(B,C)$$ Hence $(\mathcal{C},D)$ is a metric space. We garner some facts about this metric space.
Let $A⊆X$, and for each $a∈A$ let $f_a$ be a path from $f_a(0)=a$ to $f_a(1)$. The family $\{f_a\}$ is said to be uniform if for each $ε > 0$ and $r∈[0,1]$ there exists a neighborhood $U$ of $r$ in $[0,1]$ such that $f_a(U)⊆B_d(f_a(r),ε)$ for each $a∈A$. In other words, when $X^A$ is given the uniform topology, if the map $f : [0,1]→X^A$ given by $f(r)=\prod f_a(r)$ is continuous. When $B=∪f_a(1)$, we say $\{f_a\}$ is a uniform path from $A$ to $B$.
Let $σ : ℕ→I^2$ be such that $\overline{σ(ℕ)}=I^2$. Given $ε > 0$ and a function $g : ℕ→\{0,1\}$, we may define $$h_g : I^2→I$$ $$h_g(x)=\lim_{N→∞}\sum_{σ(i)∈B(x,ε)}^N \dfrac{g(i)}{N}$$ when the above limit exists for all $x∈I^2$ (also: consider weighting the sum for guaranteed convergence). When $σ$ and $ε$ are fixed, which continuous functions $f:I^2→I$ are constructible as $h_g$ for some $g∈ℕ^{\{0,1\}}$?
Let $p : I→I^2$ be a path with only finitely many self-intersection points, i.e. $|p^{-1}(x)|=1$ for all but finitely many $x∈I^2$. Choose a countable dense subset $Q⊆[0,1]$ ordered by $φ : ℕ→Q$ with the property that, for all open $U⊆[0,1]$, the sequence$$\dfrac{|\{φ(1),φ(2),...,φ(N)\}∩U|}{N}$$converges as $N→∞$ (the dyadic rationals under the usual ordering suffice). Let $V⊆[0,1]$ be a countable dense subset disjoint from $p(I)$ with ordering $ψ : ℕ→V$. When $σ : ℕ→I^2$ is defined $σ(1)=p∘φ(1)$, $σ(2)=ψ(1)$, $σ(3)=ψ(2)$, $σ(4)=p∘φ(2)$, $...$ (alternating between primes and non-primes), then $σ$ is an ordering of the countable dense subset $Q∪V$ (with the exception of finitely many points; modify without loss). Define $g : ℕ→\{0,1\}$ by $g(i)=1$ if $σ(i)∈Q$, and $g(i)=0$ otherwise. Then $h_g(x)$ denotes a well-defined function, which when the dyadic rationals as above are chosen and the limit of$$\dfrac{|\{φ(1),φ(2),...,φ(N)\}∩U|}{N}$$as $N→∞$ thus coincides with the Lebesgue measure, and measures the "length" of the segment contained the $ε$-ball about $x$. For example, when $ε=1/2$ and $p(x)=0×x$, we observe $$h_g(x,y)=\text{min }\{1,y+\sqrt{1-4x^2}/2\}-\text{max }\{0,y-\sqrt{1-4x^2}/2\}$$ for $x≤1/2$ and $h_g(x,y)=0$ elsewhere.
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