Compact cover of a Hausdorff compact space

Question

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.

Answer 1

Note: whilst typing this, Martin posted his answer. As I come to a completely different conclusion, I'd be very interested in knowing who's right!

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False. Let $M = [0,1]$ and $K_i = \{0\} \cup [\frac{1}{i},1]$.

The flaw in the proof is the assumption that the $U_j$ are increasing; ie that $U_j \subseteq U_{j+1}$. Thus the sentence "If $n_0=\max_{j=1, \dots, n} i_j$ then $x_{n_0+1} \notin \bigcup_{j=1}^n U_{i_j}$." is not true.

Answer 2

[I edited this after Andrew's comment!]

$M$ has to carry the colimit topology for the lemma to become true. Besides it is well-known and frequently used in, say, homotopy theory. Indeed, it implies for example, that homotopy groups commute with nice filtrations.

Here is a generalization to non-hausdorff spaces (which was useful in my research on algebraic geometry):

Let $X_1 \subseteq X_2 \subseteq X_3 \subseteq ...$ a sequence of closed subspaces of a topological space $X$ and let $X_\infty$ their colimit (i.e. the union with the colimit topology). Assume that $X_\infty$ is quasicompact and that for every $i < j$, such that $X_i$ is a proper subset of $X_j$, $X_j \setminus X_i$ contains a closed point of $X_j$. Then there is some $i$ such that $X_\infty = X_i$.

Proof: Assume $X_i \neq X_\infty$ for all $i$. Then you can construct $i_0 < i_1 < ...$ and $x_{i_k} \in X_{i_k} \setminus X_{i_{k+1}}$, closed in $x_{i_k}$ and thus in $X_\infty$. Now consider $D = \{a_{i_k} : k \in \mathbb{N}\}$. Then $D \cap X_i$ is a finite set of closed points of $X_\infty$ resp. $X_i$, thus closed. Therefore $D \subseteq X_\infty$ is closed. In the same way, we see that $D \setminus \{d\}$ is closed in $X_\infty$ for every $d \in D$. But then it is also closed in $D$. Hence $D$ is discrete. Since $D$ is, as a closed subset of $X_\infty$, quasicompact, we get that $D$ is finite, which is a contradiction.

EDIT: I wonder if Andrew's counterexample is so popular compared to the correct statement, which is, as I said, used frequently. :-)