In teh complete spaces, Cauchy sequences always converge to an element in the space. There are many other âexoticâ mathematical spaces that are complete and many that are not complete. In fact \(\mathbb Q\) by itself still has Cauchy sequences that do not converge (the converge to an irrational number). There are other types of sets/spaces that are not complete but still have Cauchy sequences that are not convergent. It is important here that \(\mathbb R\) is complete. Now when \(n\geq n^*\) we can just pick some arbitrary \(m\geq n^*\) such that \(|a_m-A|<\frac\epsilon2\) (since \(A\) is an accumulation point of \(S\)). Now choose a cutoff \(n^*\) such that \(n,m\geq n^*\) implies that \(|a_n-a_m|<\frac\epsilon2\) (since the sequence is assumed to be Cauchy). Now choose a cutoff \(n^*\) such that \(n\geq n^*\) implies that \(|a_n- A|0\) and choose \(n^*\) such that \(n,m\geq n^*\) implies that \(|a_n-a_m|0\) there are infinitely many \(m\) such that \(|a_m-A|<\epsilon\). Assume the sequence is convergent, \(a_n\rightarrow A\). And it is indeed a fat that every Cauchy sequence of real numbers converges toa real number. The property of âconvergenceâ depends on converging to an object that exists within the overall parent set or universe that we are considering. It will converge to an irrational number in that case. For example, \(\mathbb Q\) is an ordered field, and a Cauchy sequence of rational numbers need not converge (to a rational number). In \(\mathbb R\), a Cauchy sequence does indeed converge to a real number as we will see shortly, but in other spaces Cauchy sequences may not converge. A Cauchy sequence has its terms getting very close together eventually, but we are not necessarily given that the sequence converges. This definition looks very much like the definition of convergence, but it is slightly different. \(\blacksquare\)Äefinition 2.5.6 A sequence \(a_n\) is called a Cauchy sequence if and only if for each \(\epsilon>0\) there exists \(n^*\in\mathbb N\) such that \(|a_n-a_m|<\epsilon\), for all \(n,m\geq n^*\). This is important because being an accumulation point means that the deleted neighborhood contains infinitely many points of the set. This means that at least one of our sequences \(a_n\) or \(b_n\) is definitely not eventually constant and therefore each \(I_k\) interval does indeed contain infinitely many distinct elements of \(S\). Note that it is a fact that \(S\) contains infinitely many distinct numbers. Thus \(A\) is an accumulation point of \(S\). Therefore there are infinitely many elements of \(S\) (distinct form \(A\)) in the interval \((A-\epsilon, A+\epsilon)\). This shows that \(A-\epsilon0\), there exists at least one point \(t\) of \(S\) such that \(0<|t-s_0|<\epsilon\). S)\) implies that \(p=q \in A,\) so that \(p\) is in \(A,\) indeed.Definition 2.5.2 Let \(S\) be a set of real numbers.
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