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In his book "Riemannian Geometry" do Carmo cites the Hopf-Rinow theorem in chapter 7. (theorem 2.8). One of the equivalences there deals with the cover of the manifold using nested sequence of compact subsets. This made me wonder whether the following lemma holds:

Lemma: Let $M$ be a compact Hausdorff space, and let $K_i \subset M$ be a sequence of compact subsets such that: $K_i\subset K_{i+1}$ and $\cup_{i=1}^{\infty} K_i = M$. Then there exists an index $i_0$ such that $K_i = M$ for all $i \geq i_0$.

Here is my proof to this:

Proof: Assume that $K_i \neq M$ for all $i$, that is all $K_i$'s are proper subsets of $M$. With out loss of generality we can then assume that $K_i\subsetneq K_{i+1}$. This implies that $\forall i$ there exists $x_i$ such that $x_i\notin K_i$ but $x_i\in K_{i+1}$. Since $K_i$ is compact subset of a Hausdorff space, there exist open $U_i$ and $V_i$, such that $K_i \subset U_i$, $x_i\in V_i$ and $U_i \cap V_i = \emptyset$.

Now, if $x\in \cup_{i=1}^{\infty} U_i$ then clearly $x\in M$ since $U_i \subset M$. On the otehr hand, if $x\in M$, then $x\in K_{i_0}$ for some $i_0$, and thus it is also in $U_{i_0}$. This yields that $M= \cup_{i=0}^{\infty}U_i$.

Let us now assume that $\cup_{j=1}^n U_{i_j}=M$ is a finite cover. If $n_0 = \max_{j=1,\ldots,n}{i_j}$ then $x_{n_0+1}\notin \cup_{j=1}^nU_{i_j}$. This in turn means, that we cannot find a finite sub-cover of $M$ using the open cover $\cup_{i=1}^{\infty}U_i$. But this contradicts the compactness of $M$. This completes the proof. $\square$

Finally, here's my question. Is this lemma correct? Is my proof correct?

Thanks in advance and all the best!

Dror, Edit: As I verified with the author, he meant that the last equivalence is valid when the manifold is not compact. Thus, my false lemma, is irrelevant from the first place.

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# Compact cover of a Hausdorff compact space

In his book "Riemannian Geometry" do Carmo cites the Hopf-Rinow theorem in chapter 7. (theorem 2.8). One of the equivalences there deals with the cover of the manifold using nested sequence of compact subsets. This made me wonder whether the following lemma holds:

Lemma: Let $M$ be a compact Hausdorff space, and let $K_i \subset M$ be a sequence of compact subsets such that: $K_i\subset K_{i+1}$ and $\cup_{i=1}^{\infty} K_i = M$. Then there exists an index $i_0$ such that $K_i = M$ for all $i \geq i_0$.

Here is my proof to this:

Proof: Assume that $K_i \neq M$ for all $i$, that is all $K_i$'s are proper subsets of $M$. With out loss of generality we can then assume that $K_i\subsetneq K_{i+1}$. This implies that $\forall i$ there exists $x_i$ such that $x_i\notin K_i$ but $x_i\in K_{i+1}$. Since $K_i$ is compact subset of a Hausdorff space, there exist open $U_i$ and $V_i$, such that $K_i \subset U_i$, $x_i\in V_i$ and $U_i \cap V_i = \emptyset$.

Now, if $x\in \cup_{i=1}^{\infty} U_i$ then clearly $x\in M$ since $U_i \subset M$. On the otehr hand, if $x\in M$, then $x\in K_{i_0}$ for some $i_0$, and thus it is also in $U_{i_0}$. This yields that $M= \cup_{i=0}^{\infty}U_i$.

Let us now assume that $\cup_{j=1}^n U_{i_j}=M$ is a finite cover. If $n_0 = \max_{j=1,\ldots,n}{i_j}$ then $x_{n_0+1}\notin \cup_{j=1}^nU_{i_j}$. This in turn means, that we cannot find a finite sub-cover of $M$ using the open cover $\cup_{i=1}^{\infty}U_i$. But this contradicts the compactness of $M$. This completes the proof. $\square$

Finally, here's my question. Is this lemma correct? Is my proof correct?

Thanks in advance and all the best!