Multivariable Advanced Calculus

110 CONTINUOUS FUNCTIONSTheorem 5.8.4 Let V be a finite dimensional vector space and let ||·|| 1and ||·|| 2be two norms for V. Then these norms are equivalent which means there exist constants,δ, ∆ such that for all v ∈ Vδ ||v|| 1≤ ||v|| 2≤ ∆ ||v|| 1A set, K is sequentially compact if and only if it is closed and bounded. Also every finitedimensional normed vector space is complete. Also any closed and bounded subset of afinite dimensional normed vector space is sequentially compact.and soProof: From 5.15 and 5.16||v|| 1≤ K 1 ||v|| 2≤ K 1 K 2 ||v|| 11K 1||v|| 1≤ ||v|| 2≤ K 2 ||v|| 1.Next consider the claim that all closed and bounded sets in a normed vector spaceare sequentially compact. Let L : F n → V be defined byL (a) ≡n∑a k v kk=1where {v 1 , · · · , v n } is a basis for V . Thus L ∈ L (F n , V ) and so by Theorem 5.8.3 thisis a continuous function. Hence if K is a closed and bounded subset of V it followsL −1 (K) = F n \ L −1 ( K C) = F n \ (an open set) = a closed set.Also L −1 (K) is bounded. To see this, note that L −1 is one to one onto V and soL −1 ∈ L (V, F n ). Therefore,∣∣L −1 (v) ∣ ≤ ∣ ∣ ∣L −1∣ ∣ ∣ ||v|| ≤ ∣ ∣ ∣L −1∣ ∣ ∣ rwhere K ⊆ B (0, r) . Since K is bounded, such an r exists. Thus L −1 (K) is a closedand bounded subset of F n and is therefore sequentially compact. It follows that if{v k } ∞ k=1 ⊆ K, there is a subsequence {v k l} ∞ l=1 such that { }L −1 v kl converges to apoint, a ∈ L −1 (K) . Hence by continuity of L,v kl = L ( L −1 (v kl ) ) → La ∈ K.Conversely, suppose K is sequentially compact. I need to verify it is closed andbounded. If it is not closed, then it is missing a limit point, k 0 . Since k 0 is a limit point,there exists k n ∈ B ( k 0 , n) 1 such that kn ≠ k 0 . Therefore, {k n } has no limit point inK because k 0 /∈ K. It follows K must be closed. If K is not bounded, then you couldpick k n ∈ K such that k m /∈ B (0, m) and it follows {k k } cannot have a subsequencewhich { } converges because if k ∈ K, then for large enough m, k ∈ B (0, m/2) and so ifkkj is any subsequence, kkj /∈ B (0, m) for all but finitely many j. In other words,for any k ∈ K, it is not the limit of any subsequence. Thus K must also be bounded.Finally consider the claim about completeness. Let {v k } ∞ k=1be a Cauchy sequencein V . Since L −1 , defined above is in L (V, F n ) , it follows { }L −1 ∞v k is a Cauchyk=1sequence in F n . This follows from the inequality,∣∣∣L −1 v k − L −1 v l ≤ ∣ ∣ ∣L −1∣ ∣ ∣ ||v k − v l || .therefore, there exists a ∈ F n such that L −1 v k → a and since L is continuous,v k = L ( L −1 (v k ) ) → L (a) .