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Let $X$ be any bounded metric space and $B(X,\mathbb{R})$ - a set of all bounded functions $X\rightarrow \mathbb{R}$ which we endow with a norm $||f||=\sup_{t\in X}|f(t)|$. It is easy to verify that the map $\phi:X\rightarrow B(X,\mathbb{R})$, that sends each $x\in X$ to the function $f_x(t)=d(x,t)$ is an isometry (however, we should admit that $X$ is bounded). Also, we see that $B(X,\mathbb{R})$ is a Banach space. Hence each bounded metric space is a subspace of a Banach space.

Thus we prove that there exists a complete space $Y$ such that $X\subset Y$ and $X$ is everywhere dense in $Y$.

I want to prove this theorem without such serious suggestions on $X$. Is it possible to include each metric space into a complete space somehow (without using the standard construction with Cauchy sequences)?

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  • $\begingroup$ Do you know about Cauchy completion construction? $\endgroup$ – Moishe Kohan Oct 12 '13 at 18:13
  • $\begingroup$ Yes, I do. I want to avoid that construction $\endgroup$ – user74574 Oct 12 '13 at 18:16
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You can use the same construction as you defined for any metric space:

Define functions $f_x: X\to {\mathbb R}$ as in your question. Now, let $L(X)$ denote the space of all 1-Lipschitz functions on $X$; in particular, all functions $f_x$ belong to this set. We equip $L(X)$ with the following "metric": $$ d(f,g)=sup_{x\in X} |f(x)-g(x)|. $$ This "metric" has all the standard properties of a metric except that it is allowed to take infinite values. To remedy this, consider the subset $L_o(X)$ consisting of all $f\in L(X)$ such that $d(f, f_x)<\infty$ for some (equivalently all) $x\in X$. Now, $d$ is an ordinary metric on $L_o(X)$ and each $f_x$ belongs to $L_o(X)$. The isometric embedding $\phi: X\to f_x\in L_o(X)$ is defined in the same manner as in the question. One then verifies that $L_o(X)$ is complete (this is no different than the proof in the case when $X$ is bounded). In order to get your $Y$, take the closure of $\phi(X)$ in $L_o(X)$.

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