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Let $(X,d_{X})$ be a metric space that can be isometrically embedded in a complete metric space $(Y,d_{Y})$. What does this completion exactly mean? Is it right to say that this completion adds the "missing limits" of the Cauchy sequences in $(X,d)$ to $X$?

simp
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    That's exactly right, it adds the missing limits of Cauchy sequences. The purpose in functional analysis is seen in the fact that most of the fundamental theorems of that field require complete vector spaces. – M_B Jun 17 '17 at 22:34

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Yes, completion adds all the points to which the Cauchy sequences want to converge. This may not look like a lot at the first glance, but consider that the only way to prove that $\lim x_n$ exists without completeness is by pointing to its limit (the definition of limit requires us to know what the limit is).

This makes it impossible to construct objects (e.g., solutions of some equation) without pointing at them directly. In analysis, complex objects are usually built from simple ones by means of an infinite process that is shown to converge: consider power series, Fourier series, solution of equations by numerical methods, or the construction of von Koch snowflake and other fractals. If we can't prove convergence without already having the limit at hand, our world is sad and boring. Completeness makes it interesting to live in.