# How do I prove these facts regarding the proof of a lemma related to partitions of unity regarding local finiteness of a collection?

Consider:

Lemma $$1.4.$$ Each topological manifold is locally compact.

Proposition $$1.5.$$ Let $$X$$ be a locally compact Hausdorff space that satisfies the second countability axiom. Then there is a sequence of compact subsets of $$X: (K_j )^\infty_{j=1}$$ such that

$$K_j \subseteq K^{\text int}_{j+1} , \forall j$$ ;

$$\bigcup_{j=1}^\infty K_j=X$$.

Lemma $$1.6.$$ Let $$U \subseteq M$$ open, and $$A \subseteq U$$closed. Then there's a smooth function $$f : M \rightarrow [0, \infty)$$ so that

$$\operatorname {supp(f )} \subseteq U$$ ;

$$f(p) > 1/2 , \space \forall p \space \in A$$.

Proof: Because of Lemma 1.4, we can apply Proposition 1.5 to $$M$$. Let $$(K_j )^\infty_{j=1}$$ be as in that proposition. If $$j ≤ 0$$, then we call $$K_j:= \emptyset$$. Thenfor all $$j$$ the set

$$U_j := U \cap (K^{\text int}_{j+1} \setminus K_{j−2})$$ is open

and

$$A_j := A \cap (K_j \ \setminus K^{\text int}_{j-1}) \subseteq U_j$$ is compact

Let $$f_j$$ be as in the statement of our lemma applied to $$U_j$$ and $$A_j$$ . Call $$U_0 := M \setminus A,$$ and $$f_0 = 0$$. Then

$$\{U_j ; j = 0, 1, . . .\} \text{is a locally finite covering of }M \tag 1$$

So there is a well-defined smooth function $$f$$ on $$M$$ like that for all $$p\in M$$, $$f(p)=\sum_{j=1}^{\infty} f_j (p)$$

The support of each function $$f_j$$ lies in $$U$$, so does the support of $$f$$

$$\text{If } p \in A, \text{ then there is a }j ≥ 1 \text{ such that } p \in A_j\tag 2$$

It follows that $$f (p) ≥ f_j (p) > 1/2$$

I am trying to prove (1) and (2). I know the definition of locally finite is that a collection $$C$$ of subsets of $$M$$ is locally finite if for every $$p\in M$$ has a neighborhood that intersects at most finitely many of the sets in $$C$$. I am clueless about how to even start. I don't see how it was useful to define $$U_j$$ like that. Any help would be appreciated.

Regarding (1), the sets $$U_j$$ and $$U_k$$ are disjoint for $$k\geq j+3$$, since $$U_k\subseteq M\backslash K_{k-2}\subseteq M\backslash K_{j+1}$$ and $$U_j\subseteq K_{j+1}$$.

Therefore $$U_j$$ can only intersect $$U_{j-2}$$, $$U_{j-1}$$, $$U_{j+1}$$ and $$U_{j+2}$$, and potentially $$U_0:=M\backslash A$$, so the covering has multiplicity at most $$6$$. Moreover, the sets cover $$U$$, since if $$p\in U$$ then you can let $$j$$ be the first index for which $$p\in K_{j+1}^{\text{int}}$$, and then certainly $$p\notin K_{j}^{\text{int}}\supseteq K_{j-1}\supseteq K_{j-2},$$ so $$p\in U_j$$. Since $$U_0=M\backslash A\supseteq M\backslash U$$, the sets cover all of $$M$$.

In particular, each point $$p\in M$$ has a neighborhood intersecting at most $$6$$ members of the cover (namely, take one of the $$U_i$$'s containing $$p$$ as that neighborhood).

As for (2), for $$p\in A$$, if $$j$$ is the first index where $$p\in K_j$$ then certainly $$p\notin K^{\text{int}}_{j-1}$$, otherwise $$j$$ would not be the first such index. Therefore we get $$p\in A_j$$.

• what is "multiplicity" of a cover? Is it a definition or just your own words? Commented Oct 4, 2023 at 20:30
• @darkside It's pretty standard, just means every point is in at most $5$ members of the cover.
– M W
Commented Oct 4, 2023 at 20:33
• In the proof in my post, why do they take: $U_0:=M\setminus A$? Doesn't that contradict the definition $U_j := U \cap (K^{\text int}_{j+1} \setminus K_{j−2})$? Because for $j=0$, I have $U_0=U \cap (K^{\text int}_{1} \setminus K_{−2})=U \cap (K^{\text int}_{1} \setminus \emptyset)=U \cap K^{\text int}_{1}$ So the $U_0$ was already defined like this. Commented Oct 4, 2023 at 20:42
• Would $K^{\text int}_{j+1}\setminus K_{j-1}$ work as a a neighborhood intersecting 5 of the elements? Commented Oct 4, 2023 at 20:55
• @darkside that does look a little bit like an oversight about $U_0$, maybe their indices are off by $1$. As for your other question, it would work, since it is contained in the $U_i$'s and every element is in one of those sets, but once you know the $U_i$'s themselves have multiplicity $5$ that is already enough, you don't need to do anything further.
– M W
Commented Oct 4, 2023 at 21:01