Math and science::Topology
Compactness. Examples
- The collection of sets \( \{(n - 1, n + 1) : n \in \mathbb{Z} \} \) form an open cover of \( \mathbb{R} \). This cover has no finite subcover. So \( \mathbb{R} \) is not compact.
- Divide the interval \( [0, 1] \) like so: \( U_0 = [0, \frac{1}{2}] \), \( U_n = (2^{-n}, 1] \) for \( n \ge 1 \), then form the cover \( (U_n)_{n \ge 0} \). This is an open cover, and has many finite subcovers (e.g. ([0, 1/2), (1/2, 1], (1/4, 1]). So we can say that \( [0, 1] \) is .We can't say anything. We would need to prove that all open covers have a finite subcover to prove compactness.
- The compact subspaces of \( \mathbb{R}^n \) are the closed bounded subsets.
- Any indiscrete space is compact.
- Any finite space is compact.
- A discrete space is compact iff it is finite.
- In a normed vector space \( V \), the closed unit ball is compact iff \( V \) is finite-dimensional.
The last example alludes to the idea that compactness is a kind of finiteness condition:
| sets | finite sets |
| vector spaces | finite dimensional vector spaces |
| topological spaces | compact topological spaces |
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