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Heine-Borel Theorem

Proof by Contradiction using Interval Bisection

📜 Theorem 2.3.3 (Heine-Borel)

Let S be a closed and bounded subset of ℝ, and let {Uᵢ}ᵢ∈I be an open covering of S. Then there exists a finite subcollection that still covers S.

0Setup: Assume for Contradiction

Suppose S ⊂ [a₀, b₀] is closed and bounded, but CANNOT be covered by finitely many open sets from our covering {Uᵢ}.

We'll derive a contradiction by constructing a nested sequence of intervals.

Bisection Process: Finding the Contradictiona₀b₀[a₀, b₀]length = 1/1
Current "bad" interval
Previous intervals

🔑 Key Technique

Bisection + Pigeonhole: If the whole can't be finitely covered, at least one half can't either. Keep bisecting the "bad" half.

🎯 Why Closed Matters

The limit point c must be in S for the contradiction to work. Closedness guarantees accumulation points stay in S.

📐 Why Bounded Matters

Boundedness lets us start with a finite interval [a₀, b₀]. Unbounded sets can have open coverings with no finite subcover!

📋 Proof Logic Summary

1. Assume S cannot be finitely covered
2. Bisect repeatedly → nested intervals [aⱼ, bⱼ], each "bad"
3. Widths → 0, so sequence (cⱼ) is Cauchy → converges to c
4. S closed ⟹ c ∈ S ⟹ some Uᵢ₀ contains c
5. Small interval [aⱼ, bⱼ] fits entirely inside Uᵢ₀
6. Contradiction! [aⱼ, bⱼ] ∩ S has a finite cover (just Uᵢ₀) ∎