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I remember seeing this proof somewhere (perhaps here, but I don't remember where) that goes something like this.

Suppose $X$ is sequentially compact, and by contradiction suppose $\{U_n\}$ is a countable open cover with no finite subcover. Then for any positive integer $n$, the set $\{U_i : i \le n\}$ is not an open cover, so there exists $x_n \notin \bigcup_{i \le n} U_i$. Hence, we obtain sequence, and by sequential compactness, there exists a subsequence $x_{n_j}$ that converges to $a \in X$. However, $ a \in U_k$ for some positive integer $k$ and by construction, $x_{n_j} \notin U_k$ if $n_j \ge k$. This is a contradiction.

Doesn't this only prove every countable open cover must have a finite subcover?

  • 5
    In general, sequential compactness neither implies, nor is implied by, compactness; but for metric spaces, they are equivalent ([Wikipedia reference](http://en.wikipedia.org/wiki/Sequentially_compact_space#Examples_and_properties)).2012-06-29
  • 5
    Now that I think of it, the result was probably "Sequential compactness implies countable compactness" - not this.2012-06-29
  • 1
    Observe that for metric spaces the notions are indeed equivalent, since we can prove that sequentially compact (and compact) metric spaces are second-countable, so every open cover can be replaced with basic cover, but we have a countable basis so we have the open cover is countable - and the proof you gave follows.2012-06-29
  • 0
    See form [8 AE] at http://consequences.emich.edu/conseq.htm. $\:$2012-06-29

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