Basic Topology 4 | Compact Sets

Q: Why does openness matter in the step that pushes from x ∈ S to nearby points in S?

If x ∈ S, then x lies in some open set U_i from the cover. In the standard topology on ℝ, openness means U_i contains an epsilon-ball around x, so there exists ε > 0 with (x−ε, x+ε) ⊂ U_i. This implies that a whole right-hand interval [x, x+ε/2] sits inside U_i, so [0, x+ε/2] can be covered by the finite subcover for [0,x] plus U_i, making x+ε/2 ∈ S.

TL;DR

Compactness is defined relative to a chosen topology: the same set can be compact or not depending on which open sets are allowed.

Briefing Cornell Notes

Briefing

Compactness in topology is the rule that lets mathematicians treat certain infinite sets as if they were “almost finite” with respect to open covers. The key criterion is simple to state but powerful in use: a subset A of a topological space is compact exactly when every open cover of A can be reduced to a finite subcover. That reduction depends on the topology chosen—closed-and-bounded sets are compact in the real numbers with the standard topology, but this equivalence fails in general topological spaces.

To make the definition concrete, the discussion starts with the idea of covering a set A using open sets drawn from the topology. An open cover may involve infinitely many sets, and there is no guarantee that a finite selection will still cover everything. Finite sets behave nicely: if A has only finitely many points, any open cover can be trimmed down to finitely many members that still cover A. Compactness is precisely the extension of this “finite subcover” behavior to certain infinite sets.

Formally, given a collection of open sets {U_i} indexed by some set I, the family is an open cover of A when the union of all U_i contains A. The subset A is compact when, for every such open cover, there exists a finite subset I_0 ⊂ I such that the union over i in I_0 still covers A. The transcript emphasizes that this must hold for every possible open cover, not just for some convenient one. It also notes that other compactness characterizations—like sequence-based ones in metric spaces—are special cases of this general open-cover definition.

The real line example shows what goes wrong without compactness. Using the standard topology on ℝ, one can cover all of ℝ with open intervals (−n, n) for n ∈ ℕ. No finite collection of these intervals can cover ℝ, because any finite choice only reaches some bounded range. So ℝ is not compact.

In contrast, the unit interval [0,1] in ℝ is compact. The proof is built around an arbitrary open cover of [0,1] and a set S of points x in [0,1] such that the smaller interval [0,x] admits a finite subcover. Since [0,0] is finite, 0 lies in S. Openness then supplies “epsilon neighborhoods”: if a point is in S, the cover’s open set containing that point also contains a small interval to the right, pushing the property further. This leads to showing that the supremum of S must actually belong to S.

Finally, a contradiction seals the argument. If sup S were less than 1, openness around sup S would allow extending the finite-subcover property slightly beyond sup S, producing a number larger than sup S that still lies in S—impossible by the definition of supremum. Therefore sup S = 1, meaning [0,1] has a finite subcover for every open cover and is compact. The takeaway is that compactness is a topology-dependent “finite subcover” principle, and the unit interval in ℝ is the canonical example where infinite sets still behave finitely under open covers.

Cornell Notes

Compactness is defined through open covers: a subset A of a topological space is compact if every open cover of A has a finite subcover. This property depends on the topology, not just the underlying set. Finite sets are always compact because any cover can be reduced to finitely many open sets covering the finitely many points. In ℝ with the standard topology, ℝ itself is not compact since intervals (−n,n) cover it but no finite subcollection does. The unit interval [0,1] is compact, and the proof uses an open-cover argument with a set S of points where [0,x] has a finite subcover, then shows sup S must equal 1 via an epsilon-neighborhood step and a contradiction.

What is the exact definition of a compact subset in a general topological space?

A subset A ⊂ X is compact if for every open cover {U_i}_{i∈I} of A (meaning A ⊂ ⋃_{i∈I} U_i), there exists a finite subcollection {U_i}_{i∈I0} with I0 ⊂ I finite such that A ⊂ ⋃_{i∈I0} U_i. The requirement must hold for any open cover, not just some particular one.

Why is ℝ not compact in the standard topology?

Cover ℝ by the open intervals (−n, n) for n ∈ ℕ. This family covers every real number, but any finite subcollection uses only finitely many n values, so it is contained in (−N, N) for some largest N. That bounded interval cannot cover all of ℝ, so no finite subcover exists.

How does the proof that [0,1] is compact work at a high level?

Start with an arbitrary open cover of [0,1]. Define S = {x ∈ [0,1] : [0,x] has a finite subcover}. Show 0 ∈ S because [0,0] is finite. Then use openness: if x ∈ S, the open set covering x contains an interval to the right, so some larger point is also in S. Next show sup S ∈ S by splitting [0,sup S] into [0,x~] plus [x~,sup S] where the second part lies inside a single open set. Finally, contradiction: if sup S < 1, openness around sup S lets one extend slightly beyond sup S, producing a point larger than sup S still in S.

Why does openness matter in the step that pushes from x ∈ S to nearby points in S?

If x ∈ S, then x lies in some open set U_i from the cover. In the standard topology on ℝ, openness means U_i contains an epsilon-ball around x, so there exists ε > 0 with (x−ε, x+ε) ⊂ U_i. This implies that a whole right-hand interval [x, x+ε/2] sits inside U_i, so [0, x+ε/2] can be covered by the finite subcover for [0,x] plus U_i, making x+ε/2 ∈ S.

What contradiction arises if sup S is assumed to be less than 1?

Assume sup S < 1. The point sup S lies in some open set U_i, so there is a δ > 0 such that (sup S, sup S + δ) stays inside U_i. Then [0, sup S + δ] can be covered by the finite subcover for [0, x~] (with x~ < sup S) together with U_i covering the interval from x~ to sup S + δ. This shows sup S + δ ∈ S, contradicting that sup S is the least upper bound of S.

Review Questions

State the open-cover definition of compactness and explain what changes when passing from a finite set to an infinite one.
Give an explicit open cover of ℝ in the standard topology and explain why no finite subcover exists.
In the proof for [0,1], what is the set S, and how do epsilon-neighborhoods and the supremum argument force sup S = 1?

Key Points

1
Compactness is defined relative to a chosen topology: the same set can be compact or not depending on which open sets are allowed.
2
A subset A is compact iff every open cover of A admits a finite subcover.
3
Finite sets are automatically compact because any cover can be reduced to finitely many open sets covering the finitely many points.
4
In the standard topology on ℝ, the cover (−n,n) for n ∈ ℕ shows ℝ is not compact since no finite subcollection can cover all real numbers.
5
The unit interval [0,1] is compact, and the proof uses an arbitrary open cover plus a set S of points x where [0,x] has a finite subcover.
6
Openness in ℝ provides epsilon-neighborhoods that let the finite-subcover property extend to points slightly to the right.
7
A contradiction argument shows the supremum of S must equal 1, forcing a finite subcover for the entire interval [0,1].

Highlights

Compactness is exactly the “finite subcover for every open cover” property, and it must hold for all possible open covers.

ℝ fails compactness because bounded interval covers (−n,n) can never be reduced to finitely many sets that still reach infinity.

The [0,1] proof turns compactness into a supremum problem: define S where [0,x] has a finite subcover, then show sup S = 1.

Openness is used in a decisive way: an open set containing a point contains a whole small interval around it, enabling the extension step.

Assuming sup S < 1 forces an extension beyond the supremum, contradicting the definition of supremum. 

Topics

Compact Sets
Open Covers
Finite Subcovers
Compactness in ℝ
Compactness Proof