Every metric space can be isometrically embedded in a Banach space, so that it's a linearly independent set I'm trying to prove the following:  Every metric space can be isometrically embedded in a Banach space, so that it's a linearly independent set.
I came up with the following idea:
Let $ (X,d) $ be a metric space. We can take the vector space of all functions from $ X $ to $ \mathbb{R} $. Now we can send an element $ x \in X $ to the function which takes the value $ 1 $ on $ x $ and $ 0 $ everywhere else. This would embed $ X $ in $ \mathbb{R}^X $, so that it would be a linearly independent set and actually a basis. 
The problem seems to be finding a suitable norm, to make the embedding also an isometry. With that, we could just use the fact that every normed space can be isometrically embedded in a Banach space to get the desired result.
Any hints on constructing the metric? Or have I chosen a bad space?
 A: Hint: Perhaps try the continuous real-valued functions on $X$ with a suitable norm.
A: The metric space X embeds into a Banach space isometrically such that X is linearly independent:
Consider the vector space of all the Lipschitz functions from $X$ to $\mathbb R$, denoted by $L(X)$. Recall that a function $f:X\to\mathbb R$ is called Lipschitz if there exists constant $C$ such that
$|f(x)-f(y)| \le C d(x,y)$ for all $x,y\in X$
The infimum of $C$  is denoted by $L(f)$, which is not a norm on $L(X)$. (e.g. $L(1) = 0$).
However if we fix an element $z$ of $X$, then $L(X)$ has a norm given by
$|f| = |f(z)| + L(f)$
(to see this, both $L(f)$ and $|\text{point evaluation at} \  z|$ are seminorms. Thus the sum is a seminorm, so one needs to show that if $|f| = 0$, then $f=0$, which is easy.)
Now we wish to embed $X$ into $L(X)^*$, the continuous dual of $L(X)$. (Remark: continuous dual of a normed space is always a Banach space!).
The norm on $L(X)^*$ is the standard one:
$|F|= \sup\bigg\{|F(f)|: f\  \text{is in}\  L(X)\  \text{with}\  |f(z)|+L(f) \le 1 \bigg\}$
We first observe that point evaluations are continuous:
if $a\in X$, then $\delta_{a}$ given by $\delta_{a} (f):=f(a)$ has norm $\le d(z,a)+1$. In fact
\begin{align}|\delta_{a}|&= \sup\{|f(a)|: |f(z)|+L(f) \le 1\}\\
&\le \sup\{|f(a)-f(z)|+|f(z)|:\ |f(z)|+L(f) \le 1\}\\
&\le \sup\{L(f)\cdot d(a,z) + |f(z)|:\ |f(z)|+L(f) \le 1\}\\
&\le d(a,z)+1\end{align}
Therefore we have a well defined map $\Phi$ from $X$ to $L(X)^*$ given by $\Phi(a) = delta_a$.
Now we will show that $\Phi$ is an isometry:
Fix $a, b$. First note that
\begin{align}|\delta_a-\delta_b|&=\sup\{|f(a)-f(b)|: |f(z)|+L(f)\le1 \}\\
&\le \sup\{|f(a)-f(b)|: L(f)\le1\}\\
&\le d(a,b)\end{align}
Now, consider the function $f_a(x) = d(x,a)-d(z,a)$.
Then $f_a(z)=0$ and $L(f)\le1$. Thus $|f|\le1$. Moreover $|f_a(a)-f_a(b)| = d(a,b)$. Hence $|\delta_a-\delta_b| = d(a,b)$.
Finally given distinct $a_1,...,a_n$ in X one can show that $\delta_{a_1},...,\delta_{a_n}$ are linearly independent. If
$$c_1 \delta_{a_1} + ... c_n \delta_{a_n} = 0$$
the consider the function $g(x) = d(x,\{a_2,...,a_n\})$. $g$ is a Lipschitz function. Note that
$$(c_1 \delta_{a_1} + ... c_n \delta_{a_n}) (g) = 0$$
$$c_1 g(a_1) + c_2 g(a_2) + ...+ c_n g(a_n) = 0$$
$c_1 g(a_1) = 0$ which implies that $c_1=0$. In a similar fashion one can show that $c_2,...,c_n$ must be all 0.
This finishes the proof.
