In the proof of $\bigcup_{j=1}^n K_j=\bigcup_{j=1}^n K_j \setminus K^{\text int}_{j+1} $ , does induction allows to pass to $n=\infty$?

Question

Consider: the question that I asked here How do I prove these facts regarding the proof of a lemma related to partitions of unity regarding local finiteness of a collection?

In this case I just focusing in the following: Given that

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

• $\bigcup_{j=1}^\infty K_j=M$.

• If $j ≤ 0$, then we call $K_j:= \emptyset$

I am trying to prove explicitly that $M=\bigcup_{j=1}^\infty K_j=\bigcup_{j=1}^\infty K_j \setminus K^{\text int}_{j+1} $I was trying to do it by induction, but then I realized that what I was doing is only induction up to a certain step, but that the conclusion does not follow from induction.

$$\bigcup_{j=1}^\infty K_j\setminus K_{j-1}^{\text int} = K_1\setminus K_{0}^{\text int}\cup K_2\setminus K_{1}^{\text int} \cup ...$$ $$= K_1\setminus\emptyset \cup K_2\setminus K_{1}^{\text int} \cup ...$$ $$= K_1 \cup( K_2 \cap (K_{1}^{\text int})^C )\cup ...$$ $$= K_1 \cup (K_2 \cap \overline{(K_{1}^C)}) \cup ...$$ $$= [(K_1 \cup K_2)\cap (K_1 \cup \overline{(K_{1}^C)}))] \cup...$$ $$= [(K_1 \cup K_2)\cap (\overline{K_1} \cup \overline{(K_{1}^C)}))] \cup...$$ $$= [(K_1 \cup K_2)\cap (\overline{K_1 \cup (K_{1}^C)})] \cup...$$ $$=[(K_1 \cup K_2)\cap (\overline{M})] \cup...$$ $$=[(K_1 \cup K_2)\cap M] \cup...$$ $$=[(K_1 \cup K_2)] \cup...$$ This is actually induction, I know I should have written the basis step and then the inductive step separately, but I realized after I had typed it all, sorry.

So this is induction corresponding to proving that:

$\bigcup_{j=1}^n K_j\setminus K_{j-1}^{\text int}=\bigcup_{j=1}^n K_j $

but I wanted to prove that

$\bigcup_{j=1}^\infty K_j\setminus K_{j-1}^{\text int}=\bigcup_{j=1}^\infty K_j $ and I don't think induction allows me to do this passage to the limit, does it? If yes, why?If not, how do I justify it then?

Additionaly, are the topological manipulations regarding the interior, closure and complement correct? I have used that a compact set is closedm so it coincides with its closure, because we're in a manifolds, so it is Hausdorff, and that the closure of a union is the union of the closures

Answer 1

Let's say $x \in M$, then $x$ must be in one of the $K_n$. As the collection $\{K_1, K_2, ..., K_n \}$ is finite, we can find the minimum $m$ such that $x \in K_m$ and $x \notin K_{m-1}$ (by convenience, assume $K_0 = \emptyset$). Therefore, $x \in K_{m} \setminus K_{m-1} \subseteq K_{m} \setminus K_{m-1}^{int}$