Compactness · Definition

Sequential compactness

topologycompactnesssequences

Before you start you need to know

It translates compactness into the language of sequences, the most convenient one for working in metric spaces and in analysis.

Definition

A metric space \(X\) is sequentially compact if every sequence \((x_n)\) in \(X\) has a subsequence \((x_{n_k})\) converging to a point \(x\in X\).

\[\forall (x_n)\subseteq X\ \ \exists\, n_1

In \(\mathbb{R}\) the property is the Bolzano-Weierstrass theorem: every bounded sequence has a convergent subsequence. That is why \([a,b]\) is sequentially compact, and by Heine-Borel the same holds for every closed and bounded subset of \(\mathbb{R}^n\). What matters is that the limit must live in the space: \(x_n=1/n\) has convergent subsequences in \(\mathbb{R}\), but all tend to \(0\), which does not belong to \((0,1)\).

Counterexamples almost always have the same shape. \(\mathbb{R}\) fails the property because \(x_n=n\) has no convergent subsequence, since every subsequence is unbounded. An infinite set with the discrete distance fails too: a sequence of distinct points would have convergent subsequences only if they were eventually constant, and they are not.

The central result is that, in a metric space, compact and sequentially compact are equivalent. If \(X\) is compact and a sequence had no convergent subsequence, every point would have a ball containing terms for only finitely many indices; finitely many balls would cover \(X\), leaving only finitely many indices in total, an absurdity. The converse uses the Lebesgue number: a sequentially compact space is totally bounded, and every open cover has a Lebesgue \(\delta\), from which a finite subcover follows.

In general topological spaces the two notions come apart: there are compact spaces that are not sequentially compact and sequentially compact spaces that are not compact. That is why the definition is given here for metric spaces, where it agrees with compactness and lets you argue with sequences instead of covers.

Think about it

Is the space \([0,1]\cap\mathbb{Q}\) with the usual distance sequentially compact?

No: a sequence of rationals in \([0,1]\) converging to \(1/\sqrt2\) has all its subsequences converging to that irrational number, which is not in the space.

Why must the limit in the definition lie in \(X\)?

Because if it were enough for the subsequence to converge in a larger space, \((0,1)\) would be «compact» thanks to \(1/n\to0\); the condition demands that the space contain its own limits.

Exercises 10

  1. Which of these spaces, with the usual distance, is sequentially compact?
  2. Why is \(\mathbb{R}\) not sequentially compact?
  3. To prove that \((0,1)\) is not sequentially compact, which sequence works?
  4. In \(X=(1,2]\) with the usual distance, does the sequence \(x_n=1+\tfrac1n\) have a subsequence converging in \(X\)?
  5. Which of these subsets of \(\mathbb{R}^2\), with the Euclidean distance, is sequentially compact?
  6. Let \(x_n=\cos\!\big(\tfrac{n\pi}{2}\big)\) in \([-1,1]\). Which values are limits of some subsequence?
  7. Let \(x_n=(-1)^n\,\dfrac{n}{n+1}\) in \([-1,1]\). To which values do its convergent subsequences converge?
  8. For a metric space \(X\), which statement is true?
  9. Let \(K\) be a sequentially compact metric space and \(F\subseteq K\) closed. Is \(F\) sequentially compact?
  10. Let \(X\) be infinite with the discrete distance (\(d=1\) between distinct points). Is it sequentially compact?

References

  • James R. Munkres, Topology, 2nd edition, Pearson, 2000. §28 (Teorema 28.2)
  • Bert Mendelson, Introduction to Topology, 3rd edition, Dover Publications, 1990. cap. 2