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B) of R2 is not a complete metric space. one can’t infer whether a metric space is complete just by looking at the underlying topological space. Since is a complete space, the sequence has a limit. The Completion of a Metric Space Let (X;d) be a metric space. The resulting space will be denoted by Xand will be called the completion of … Let (X, d) be a metric space. Let be a Cauchy sequence in the sequence of real numbers is a Cauchy sequence (check it!). Completion of a Metric Space Definition. Already know: with the usual metric is a complete space. Let (X;d X) be a complete metric space and Y be a subset of X:Then (Y;d Y) is complete if and only if Y is a closed subset of X: Proof. A completion of a metric space (X, d) is a pair consisting of a complete metric space (X *, d *) and an isometry ϕ: X → X * such that ϕ [X] is dense in X *. For example, consider the real line \$\mathbb{R}\$ and the open unit interval \$(-1,1)\$, each with the usual metric. Completeness is not a topological property, i.e. Theorem 1. is complete if it’s complete as a metric space, i.e., if all Cauchy sequences converge to elements of the n.v.s. This proposition allows us to construct many examples of metric spaces which are not complete. Proof. Every metric space has a completion. with the uniform metric is complete. A metric space (X, d) is said to be complete if every Cauchy sequence in X converges. A metric space is called complete if every Cauchy sequence converges to a limit. The goal of these notes is to construct a complete metric space which contains X as a subspace and which is the \smallest" space with respect to these two properties. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Completeness is not a topological property, that is, there are metric spaces which are homeomorphic as topological spaces, one being complete and the other not. Proof. Hence, we will have to make some adjustments to this initial construction, which we shall undertake in the following sections. Therefore our de nition of a complete metric space applies to normed vector spaces: an n.v.s. The hope with this initial construction is that (C[E];D) is a complete metric space, but, as will be seen in part (v) of Exercise 1.2, Dfails to even be a metric. This is left to the reader as an exercise. Proposition 1.1. Recall that every normed vector space is a metric space, with the metric d(x;x0) = kx x0k. Theorem. Denote by C [X] the collection of all Cauchy sequences in X. 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