Functional Analysis 16 | Compact Sets [dark version]

TL;DR

Compactness in metric spaces is defined sequentially: every sequence has a convergent subsequence whose limit remains inside the set.

Briefing Cornell Notes

Briefing

Compactness in functional analysis is often summarized as “closed and bounded,” but that shortcut only works in familiar settings like 2 and 3 with the standard metric. The core takeaway here is a more general, metric-space definition: a set is compact exactly when every sequence inside it has a subsequence that converges to a point still inside the set. This sequential viewpoint turns compactness into a practical test for infinite sets: no matter how points are chosen endlessly, some infinite “tail” must settle down to a limit that stays within the set.

In 2 (and similarly 3), two geometric restrictions already hint at compactness. A closed set prevents “escape through the boundary”: if a sequence stays in the set and converges, its limit must also lie in the set. A bounded set prevents “escape to infinity”: all points remain within some finite distance from each other (equivalently, the set fits inside a large enough ball). Together, these two properties match compactness in Euclidean spaces, and the sequential definition reproduces the familiar BolzanoWeierstrass theorem: every bounded sequence in 2 has a convergent subsequence.

The transcript then stresses why the metric matters. Using the unit interval b0,1d as an example, it is compact under the usual Euclidean metric because sequences like b1/nd have convergent subsequences whose limits remain in b0,1d. But switch to the discrete metric, where any two distinct points have distance 1, and compactness fails. The sequence b1/nd stays inside b0,1d, yet no subsequence can converge: in the discrete metric, convergence would require distances to shrink to 0, which never happens unless the sequence becomes eventually constant. That single constructed sequence is enough to show the unit interval is not compact in the discrete metric.

Finally, a general proposition is proved for metric spaces: if a set is compact, then it must be closed and bounded. The “closed” part follows by taking any convergent sequence in the set; compactness guarantees a subsequence converging to a limit that lies in the set, forcing the original limit to belong to the set as well. For “boundedness,” the argument proceeds by contradiction: if the set were unbounded, one could build a sequence whose points run farther and farther away from a chosen reference point. Such a sequence cannot have any convergent subsequence, contradicting compactness. The result is one-way: closed and bounded are necessary in metric spaces, but not automatically sufficient unless one is in the Euclidean/standard-metric setting.

The practical message is clear: compactness is a sequential constraint on infinite behavior, and in metric spaces it always enforces both closedness and boundedness. That framework sets up later extensions from sets to operators in functional analysis.

Cornell Notes

Compactness in metric spaces is defined sequentially: a set is compact if every sequence in the set has a subsequence that converges to a limit point still inside the set. In Euclidean spaces (like 2 and 3 with the standard metric), this matches the familiar criterion that compact sets are exactly those that are closed and bounded. The metric can change the outcome: the unit interval b0,1d is compact with the usual Euclidean metric but not compact with the discrete metric, because sequences like b1/nd have no convergent subsequence under the discrete distance rule. A general proposition holds in any metric space: compactness implies the set is closed and bounded, though the reverse implication may fail outside the Euclidean setting.

What does “compact” mean in a metric space when phrased in terms of sequences?

A subset A of a metric space is compact (sequentially compact) if for every sequence (x_n) with x_n A, there exists a subsequence (x_{n_k}) that converges to some limit point A. The key requirement is that the limit of the convergent subsequence must still lie inside A.

Why do closedness and boundedness appear as the “right” restrictions in 2 with the standard metric?

Closedness prevents limits from landing outside the set: if a sequence in A converges, its limit must belong to A. Boundedness prevents escape to infinity: all points of A stay within some finite distance scale. In 2/3 with the standard metric, these two restrictions align with the sequential compactness behavior captured by BolzanoWeierstrass.

How can the same set be compact under one metric and not compact under another?

The unit interval b0,1d is compact with the Euclidean metric, but not with the discrete metric. Under the discrete metric, any two distinct points have distance 1, so a sequence can only converge if it becomes eventually constant. The sequence x_n = 1/n stays in b0,1d but never becomes constant, so it has no convergent subsequence; that failure shows non-compactness.

What does compactness force about a set in any metric space, according to the proposition?

If A is compact in a metric space, then A must be closed and bounded. Closedness is shown by taking a convergent sequence in A and using compactness to ensure the limit lies in A. Boundedness is shown by contradiction: if A were unbounded, one can construct a sequence that runs arbitrarily far from a reference point, preventing any convergent subsequence, which contradicts compactness.

Why does unboundedness contradict compactness in the proof?

If A is unbounded, for a fixed point a A one can choose points x_n A with d(a, x_n) > n. Any subsequence still keeps distances from a growing without bound, so for any candidate limit point B, the triangle inequality prevents d(B, x_{n_k}) from going to 0. That means no subsequence can converge, contradicting compactness.

Review Questions

State the sequential definition of compactness in a metric space and explain what must happen to the limit point.
Give an example of a set that is compact under one metric but not under another, and identify the sequence that breaks compactness.
In a metric space, what properties are guaranteed by compactness, and which direction does the implication go?

Key Points

1
Compactness in metric spaces is defined sequentially: every sequence has a convergent subsequence whose limit remains inside the set.
2
In 2 and 3 with the standard metric, compactness matches the closed-and-bounded criterion (via BolzanoWeierstrass).
3
The metric choice can change compactness: b0,1d is compact with the Euclidean metric but not with the discrete metric.
4
Under the discrete metric, convergence requires a sequence to be eventually constant, so sequences like 1/n have no convergent subsequence.
5
In any metric space, compactness implies the set is closed: limits of convergent sequences from the set must stay in the set.
6
In any metric space, compactness implies the set is bounded: unbounded sets allow sequences that cannot have convergent subsequences.
7
The relationship “closed and bounded compact” is not automatic in general metric spaces; compactness is the stronger condition.

Highlights

Compactness becomes a sequence-and-subsequence condition: infinite choices must contain an infinite subsequence that settles to a limit inside the set.

The unit interval b0,1d is compact with the Euclidean metric but fails under the discrete metric because 1/n never yields a convergent subsequence.

A general theorem for metric spaces: compact  closed and bounded (one-way implication).

Unboundedness can be turned into a “runaway” sequence whose distances prevent any subsequence from converging.

Topics

Compact Sets
Metric Spaces
Sequential Compactness
Closed and Bounded
Discrete Metric