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


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


1 Answer 1


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 $p\in A_j$.

  • $\begingroup$ what is "multiplicity" of a cover? Is it a definition or just your own words? $\endgroup$
    – darkside
    Commented Oct 4, 2023 at 20:30
  • $\begingroup$ @darkside It's pretty standard, just means every point is in at most $5$ members of the cover. $\endgroup$
    – M W
    Commented Oct 4, 2023 at 20:33
  • $\begingroup$ 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. $\endgroup$
    – darkside
    Commented Oct 4, 2023 at 20:42
  • $\begingroup$ Would $K^{\text int}_{j+1}\setminus K_{j-1}$ work as a a neighborhood intersecting 5 of the elements? $\endgroup$
    – darkside
    Commented Oct 4, 2023 at 20:55
  • $\begingroup$ @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. $\endgroup$
    – M W
    Commented Oct 4, 2023 at 21:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .