Theorem 2.3-2 in _Introductory Functional Analysis With Applications_ by Erwine Kryszeg Here's the statement of Theorem 2.3-2 in the book mentioned above: 
Let $(X,||\cdot||)$ be a normed space. Then there is a Banach space $\hat{X}$ and an isometry $A \colon X \to W$, where $W = A(X)$, and $W$ is dense in $\hat{X}$. This space $\hat{X}$ is unique except for isometries. 
Now in the proof of the theorem, Kryszeg gives the norm on $\hat{X}$ as follows: 
For every $\hat{y} = Ax \in W$, we have $||\hat{y}||_1 = ||x||$. 
And, for every $\hat{x} \in \hat{X}$, we have $||\hat{x}||_2 \colon= \hat{d}(\hat{x},\hat{0})$. 
Now when it comes to verifying the the scalar-multiple and the triangle inequality for the norm $||\cdot||_2$, Kryszeg states that these properties follow from those for $||\cdot||_1$ by a limit process. 
How to explicitly verify these properties? 
 A: The usual definition of completion of a normed space is obtained by considering equivalence classes $[\{ x_{n} \}]$ of Cauchy sequences, grouped according to the equivalence relation that
$$\{ x_{n} \} \sim \{ y_{n} \} \iff \lim_{n}(x_{n}-y_{n})=0.$$
This $\sim$ is an equivalence relation because it is (a) reflexive (b) symmetric and (c) transitive:


*

*(a) reflexive: $\{ x_{n} \} \sim \{ x_{n} \}$

*(b) symmetric: $\{ x_{n} \} \sim \{ y_{n} \}$ iff $\{ y_{n} \} \sim \{ x_{n} \}$ and

*(c) transitive: $\{ x_{n} \} \sim \{ y_{n} \}$ and $\{ y_{n} \} \sim \{ z_{n} \}$ $\implies$ $\{ x_{n} \} \sim \{ z_{n} \}$


An equivalence relation partitions the set of Cauchy sequences into classes (they're not called sets because sets of sets can get you in paradoxes,) known as equivalence classes, and denoted by $[\{ x_n\}]$. This equivalence class $[\{x_{n}\}]$ denotes the class of all Cauchy sequences which are equivalent to each other and to a given Cauchy sequence $\{ x_{n} \}$. Two equivalence classes are either disjoint or they're the same. That is $[\{ x_{n} \}]=[\{ y_{n}\}]$ iff $\{ x_{n} \} \sim \{ y_{n}\}$. That's why an equivalence class requires only one representative element to define the equivalence class, and, hence, the notation $[\{ x_{n} \}]$ to designate an equivalence class.
This is a natural generalization of the construction to complete the rational numbers to the reals, where one defines a real number to an equivalence class of Cauchy sequences of rational numbers. There, for example, $\sqrt{2} =[\{ 1, 1.4,1.41, 1.414, 1.4142,\cdots\} ]$, which has a clear and natural meaning in terms of decimal expansions. Any rational number such as $2.1$ is embedded in this completion through the equivalence class $[\{2.1,2.1,2.1,\cdots\}]$ containing the constant sequence. Any Cauchy sequence $\{ x_{n} \}$ of positive rational numbers for which $[ \{ x_{n}^{2}\}]=2 =[\{2,2,2,\cdots\}]$ will give the same equivalence class designated as $\sqrt{2}$. This type of construction is one of the crowning achievements of 19th century Mathematics, and this construction ultimately led to a logically consistent definition of the real numbers as a complete space; and this particular construction of the completion--unlike most others--can also be used to complete a broad class of spaces where a metric is defined.
Back to the normed space $X$. Define $\hat{X}$ to be the equivalence classes of Cauchy sequences from $X$. If $[ \{ x_{n} \} ]$ is an equivalence class, then regardless of which representative sequence $\{ y_{n} \}$ that you choose in that equivalence class, the following exists and equals the same number: $\lim_{n}\|y_{n}\|$. The reason this limit exists is that the Cauchy sequence $\{ y_{n}\}$ is mapped to a Cauchy sequence of real numbers $\{\;\|y_{n}\|\;\}$, and $\{\;\|x_{n}\|\;\}\sim \{\;\|y_{n}\|\;\}$ as Cauchy sequences of real numbers whenever $\{ x_{n} \} \sim \{ y_{n}\}$ as Cauchy sequences of vectors. This is due to the reverse triangle inequality:
$$
          |\;\|x_{n}\|-\|y_{n}\|\;| \le \|x_{n}-y_{n}\|.
$$
So that defines a norm on an equivalence class which is independent of the sequence $\{ x_{n} \}$ chosen from the equivalence class:
$$
               \|\;[\{ x_{n} \}]\;\|=\lim_{n}\|x_{n}\|.
$$
There is no confusion in using the same norm symbol because the objects are clearly different. There are natural vector operations defined on these equivalence classes of Cauchy sequences:
$$
                    [\{ x_{n} \}]+[\{y_{n}\}] = [\{ x_{n}+y_{n}\}],\\
                   \alpha [\{ x_{n}\}] = [\{ \alpha x_{n}\}].
$$
These operations make sense because $\{ x_{n} \} \sim \{ x_{n} '\}$ and $\{ y_{n} \} \sim \{ y_{n}'\}$ together imply that $\{ x_{n}+y_{n} \} \sim \{ x_{n}'+y_{n}'\}$ and $\{\alpha x_{n} \}\sim \{ \alpha x_{n}'\}$. This defines a normed linear space $\hat{X}$. Using a tedious Cantor diagonalization method, one can show that $\hat{X}$ is a complete normed linear space.
In the same way that the rational numbers $\mathbb{Q}$ are naturally embedded in $\mathbb{R}=\hat{\mathbb{Q}}$ through the correspondence $q \mapsto [\{ q,q,q,q,q,\cdots\}]$, our vector space $X$ has a natural embedding in this new space $\hat{X}$ through the map $x \mapsto [\{ x,x,x,x,\cdots\}]$, meaning that $x$ is mapped to the equivalence class containing the constant sequence $\{ x,x,x,\cdots\}$. This correspondence is linear and isometric, and $X$ is dense in $\hat{X}$ because a Cauchy sequence $\{ x_{n} \} \subset X$ is mapped to a sequence $\{ \hat{x_{n}}\}\subset \hat{X}$ which can be shown to converge to $[\{ x_{n}\}] \in \hat{X}$.
A completion $\hat{X}$ of $X$ is unique. Of course, by completion one means a complete normed linear space $\hat{X}$ and an isometric linear map $x\in X\mapsto \hat{x} \in \hat{X}$ such that the image of $X$ in $\hat{X}$ is dense. Every $\hat{x}\in \hat{X}$ is easily identified with an equivalence class of Cauchy sequences in $X$. And that's where Kreyszeg is getting the norm. For a general $\hat{y} \in \hat{X}$, there exists a sequence $\{ x_{n} \}\subset X$ such that $\{ \hat{x_{n}}\}$ converges to $\hat{y}$, and you can define $\|\hat{y}\|_{\hat{X}}=\lim_{n}\|x_{n}\|_{X}$; this norm will be the same for all equivalent Cauchy sequences $\{ x_{n} \}$ because, basically, $\hat{y} = [\{ x_{n}\}]$. In other words, the standard construction of the completion is unique.
