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Let $X$ be a normed space. We say that $X$ is complete if every Cauchy sequence in $X$ converges to an element of $X$ in norm. Now, in proofs of completeness, we start with $\{x_n \}$ Cauchy, and it seems we should first exhibit an element $x$ of $X$ and show that $\{x_n \}$ converges to $x$ in norm. But, I read a proof where first, they constructed the limiting element $x$, and then showed that $\{ x_n\}$ tend to $x$ under the norm, and used that convergence result to argue that $x$ was actually a member of the space $X$. I'm not sure why I find this argument fishy. I feel you can't say something converges in norm if the limiting element doesn't even exist in your space until after it converges...

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If you know $X$ is complete that surely $x$ must belong to $X$, otherwise it is easy to construct a counterexample e.g. by looking at rational numbers converging to some none rational number. –  AD. Mar 8 '12 at 7:37
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I reference to the proof you mention might help. –  AD. Mar 8 '12 at 7:38
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up vote 1 down vote accepted

The kind of proof you describe is common in spaces of functions (or sequences, which are a special sort of functions). Typically, a function space $X$ contains only functions $f: E\to\mathbb R$ (or another field) with some specific properties. Given a sequence of functions $f_n$, it is natural to first exhibit a function to which they converge and then prove it has the properties that qualify it is a member of $X$.

For example: given a sequence of continuous function $f_n$ that is Cauchy in norm, one argues that for each $x\in E$ the limit $\lim_{n\to\infty} f_n(x)$ exists. Defining $f(x)$ to be this limit, we get a function $f:E\to\mathbb R$. Since we don't know yet that $f\in X$, we are not ready to argue that $f_n\to f$ in $X$. What we have so far is some function $f$ to which $f_n$ converge in some sense. Having a limit in some sense is better than nothing: it gives us a base to build on. A $3\epsilon$-argument is used to prove that $f$ actually is an element of $X$. After that, seeing that $f_n\to f$ in the norm of $X$ is easy.

In a nutshell: even though the ultimate goal is convergence in $X$, our proof may involve objects outside of $X$, and notions of convergence other than convergence in $X$.

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