By Sloughter D.

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Hence the only possible subsequential limits of {xn }∞ n=1 would be −∞ and +∞, contradicting our assumptions. Thus K must be bounded. Now suppose {xn }n∈I is a convergent sequence with xn ∈ K for all n ∈ I. If L = lim xn , then L is the only subsequential limit of {xn }n∈I . Hence, by the n→∞ assumptions of the proposition, L ∈ K. Hence K is closed. Since K is both closed and bounded, it is compact. D. 5. Show that a set K ⊂ R is compact if and only if every infinite subset of K has a limit point in K.

9) β∈B Then {Uβ : β ∈ B} ∪ {Uα } ∈ O is a finite subcover of I. Thus I is compact. D. 2. If K is a closed, bounded subset of R, then K is compact. Proof. Since K is bounded, there exist finite real numbers a and b such that K ⊂ [a, b]. Let {Uα : α ∈ A} be an open cover of K. Let V = R \ K. 11) is an open cover of [a, b]. Since [a, b] is compact, there exists a finite subcover of this cover. This subcover is either of the form {Uβ : β ∈ B} or {Uβ : β ∈ B} ∪ {V } for some B ⊂ A. 12) β∈B in the latter case, we have K ⊂ [a, b] \ V ⊂ Uβ .

8) Moreover, there exists {Uβ : β ∈ B} ∈ O such that a, b − 2 ⊂ Uβ . 9) β∈B Then {Uβ : β ∈ B} ∪ {Uα } ∈ O is a finite subcover of I. Thus I is compact. D. 2. If K is a closed, bounded subset of R, then K is compact. Proof. Since K is bounded, there exist finite real numbers a and b such that K ⊂ [a, b]. Let {Uα : α ∈ A} be an open cover of K. Let V = R \ K. 11) is an open cover of [a, b]. Since [a, b] is compact, there exists a finite subcover of this cover. This subcover is either of the form {Uβ : β ∈ B} or {Uβ : β ∈ B} ∪ {V } for some B ⊂ A.