![]() ![]() Then infinitely many sequence terms are outside the neighborhood, and so the sequence can't converge to the original point. Theorem 1.12 Cauchy implies bounded In a metric space, every Cauchy sequence is bounded. rem, the Cauchy sequence (xn) has a convergent subsequence (xnk ). If a Cauchy sequence has a convergent subsequence, then the sequence itself is convergent with the same limit. Then by Theorem 5.3 the subsequence y n k f(x n k) converges to the limit f( ) 2f(K). ![]() If you want try proving the following: Everey bounded sequence of real numbers has a converging subsequence Rightarrow every Cauchy sequence of real numbers converges. Edit: Obviously there are other definitions of the real numbers. ![]() Theorem 1.11 Convergent implies Cauchy In a metric space, every convergent sequence is a Cauchy sequence. Every convergent sequence is a Cauchy sequence. Since Kis compact, the sequence (x n) has a subsequence (x n k) which con-verges to a limit 2K. The notion of convergence does not always make sense. Proof: Let fxng x, let 0, let n be such that n > n ) d(xn x) < 2, and let m n > n.We know (as a consequence of the Eberlein-Šmulian theorem) that any bounded sequence, $\^\prime$, so $\lim_n \varphi(x_n) = \Psi(\varphi)(y)$ for all $\varphi \in X^\prime$.S)\) implies that \(p=q \in A,\) so that \(p\) is in \(A,\) indeed. For part 1, if there were a subsequence that didn't converge to the same limit, then we could find a neighborhood around the original point such that infinitely-many subsequence terms were outside the neighborhood. Cauchy if, given any > 0there exists an integer N such that d(x m,x n)< for all m,n N. Unlike a complete metric space, Bourbaki completeness is defined as every BourbakiCauchy sequence in. If every sequence of points of X contains a convergent subsequence, then it is called a sequentially compact space. Remark: Every convergent sequence is Cauchy. ![]()
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