Sigma-Locally Finite Cover and Countable Locally Finite Cover have Common Locally Finite Refinement

Theorem
Let $T = \struct{X, \tau}$ be a topological space.

Let $\SS = \ds \bigcup_{n = 0}^\infty \SS_n$ be a $\sigma$-locally finite cover of $X$, where each $\SS_n$ is locally finite for all $n \in \N$.

Let $\CC = \set{C_n : n \in \N}$ be a countable locally finite cover of $X$.

Then:
 * there exists a common locally finite refinement $\AA$ of both $\SS$ and $\CC$

Proof
Let:
 * $\AA = \set{C_n \cap S : n \in \N, S \in \SS_n}$

$\AA$ is a Cover of $X$
Let $x \in X$.

By definition of a cover:
 * $\exists n \in \N : x \in C_n$

and
 * $\exists S \in \SS : x \in S$

By definition of set union:
 * $\exists n \in \N : S \in SS_n$

By definition of set intersection:
 * $x \in C_n \cap S$

By definition of $\AA$:
 * $C_n \cap S \in \AA$

It follows that $\AA$ is a cover by definition.

$\AA$ is a Refinement of $\SS$ and $\CC$
Let $A \in \AA$.

By definition of $\AA$:
 * $\exists n \in \N, S \in SS_n : A = C_n \cap S$

From Intersection is Subset:
 * $A \subseteq C_n$ and $A \subseteq S$

Hence $\AA$ is a refinement of both $\SS$ and $\CC$ by definition.

$\AA$ is Locally Finite
Let $x \in X$.

By definition of locally finite:
 * $\exists U \in \tau : x \in U : \set{C_n \in \CC : C_n \cap U \ne \O}$ is finite

From Subset of Naturals is Finite iff Bounded:
 * $N = \set{n \in \N : C_n \cap U \ne \O}$ is bounded

Hence $N$ has a greatest element $m$.

By definition of locally finite:
 * $\forall n \le m : \exists V_n \in \tau : x \in V_n : \set{S \in \SS_n : S \cap V_n \ne \O}$ is finite

Let:
 * $W = U \cap \ds \bigcap_{n \le m} V_n$

By :
 * $W \in \tau$

Let $A \in \AA : A \cap W \ne \O$.

Hence there exists $n \in \N, S \in \SS_n:$
 * $A = C_n \cap S$

From Subsets of Disjoint Sets are Disjoint:
 * $C_n \cap U \ne \O$

Hence $n \le m$.

From Subsets of Disjoint Sets are Disjoint:
 * $S \cap V_n \ne \O$

We have:

From Finite Union of Finite Sets is Finite:
 * $\ds \bigcup_{n \le m} \set{C_n \cap S : S \in \SS_n, S \cap V_n \ne \O}$ is finite.

From Subset of Finite Set is Finite:
 * $\set{A \in \AA : A \cap W \ne \O}$ is finite

Hence $\AA$ is locally finite by definition.