Question

6. Suppose K1 and K2 are compact. Why is K1 ∪ K2 necessarily also compact?

(a) Write a proof of this using the sequential definition.

(b) Write a proof of this using the “closed and bounded” definition.

(c) Write a proof of this using open covers and subcovers.

Answer 1

Given K₁ , K₂ are compact set .

(a). Let (x_n) be a sequence in K₁ K₂ .

at least one of K₁ and K₂ contains infinite number of elements of the sequence (x_n) .

If K₁ contains infinite number of points then it has a covergent subsequence and if K₂ contains infinite number of points then it has a convergent subsequence on K₂ and so in K₁ K₂ .

So in either case (x_n) has a convergent subsequence in K₁ K₂ .

Hence K₁ K₂ is compact .

(b) . As K₁ and K₂ both are compact set so they are closed and bounded .

Now union of two closed sed is closed so K₁ K₂ is closed .

Also union of two bounded sed is bounded so K₁ K₂ is bounded.

As K₁ K₂ is closed as well as bonded so it is compct .

(c). Let U₁ , U₂ , U₃ ,........ be a open cover for K₁ K₂ .

As K₁ K₁ K₂ so U₁ , U₂ , U₃, ....... also be a open cover for K₁ and as K₁ is compact this cover has a finite subcover say V₁, V₂,....,V_n ( here each V_i is some U_j )

Similarly as K₂ K₁ K₂ so U₁ , U₂ , U₃, ....... also be a open cover for K₂ and as K₂ is compact this cover has a finite subcover say W₁,W₂,......,W_m

V₁, V₂ , V₃ ,....,V_n, W₁ , W₂,.....,W_m be a finite subcover for K₁ K₂ .

So the cover U₁ , U_{2 , .......} has a finite subcover .

Hence K₁ K₂ is compact .

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