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.


2 Answers 2


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!

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.

  • $\begingroup$ Andrew, your counterexample is certainly sound. Of course, there may be some reasonable natural extra conditions you could put on the $K_i$ to rescue the conclusion. $\endgroup$ Commented Aug 2, 2010 at 9:44
  • $\begingroup$ +1! Sometimes I happen to believe this lemma too, till I remember of the counterexample. $\endgroup$ Commented Aug 2, 2010 at 9:47
  • $\begingroup$ What if one also requires that the $K_i$ are $connected$? $\endgroup$ Commented Aug 2, 2010 at 9:51
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    $\begingroup$ For connected $K_i$, one could adapt Andrew's example to $M$ being the unit circle. $\endgroup$ Commented Aug 2, 2010 at 9:54
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    $\begingroup$ Which lemma? If you mean the fact that compact Hausdorff spaces are normal, then yes, you can apply that lemma. But that application is within $K_{i+1}$, and so $U_i$ is open relative to $K_{i+1}$. When passing to $M$, we need to enlarge $U_i$ to make it relative to $M$, and this destroys the condition $U_i \subseteq K_{i+1}$. Try writing down the $U_i$ for the example I gave; try both as relative to $K_{i+1}$ and as subsets of $M$ and you'll see what I mean. The thing to focus on is the $0$. $\endgroup$ Commented Aug 2, 2010 at 10:39

[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. :-)

  • $\begingroup$ Thanks for the quick answer! Could you add a reference where this lemma is cited? I couldn't find one. $\endgroup$ Commented Aug 2, 2010 at 9:26
  • $\begingroup$ I've reached the opposite conclusion! $\endgroup$ Commented Aug 2, 2010 at 9:33
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    $\begingroup$ If the $X_i$ are the sets in Andrew's counterexample, the colimit of the $X_i$ is the disjoint union of $\{pt\}$ and $[0,\infty)$ an isn't compact. $\endgroup$ Commented Aug 2, 2010 at 9:43
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    $\begingroup$ The union of an increasing sequence of subsets of a topological space need not be homeomorphic to the direct limit of that sequence. $\endgroup$ Commented Aug 2, 2010 at 9:57
  • $\begingroup$ Very nice! This is closely related to the standard proof that compact subsets of CW complexes always lie inside a finite subcomplex (see the Appendix to Hatcher's book, for example). $\endgroup$
    – Dan Ramras
    Commented Aug 2, 2010 at 18:31

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