# 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?

• The set you constructed is NOT a basis for $\Bbb R^X$. Oct 29, 2014 at 18:52
• Your embedding does not use the metric $d$. Try to embed $X$ into a space of functions by associating to every $x$ a function built from $x$ and $d$. Oct 29, 2014 at 18:54
• Probably useful: en.wikipedia.org/wiki/Kuratowski_embedding Oct 29, 2014 at 18:57
• I've been reading about Kuratowsi embeddings, but can they/ something similar be used to densely embed any seperable metric space into a Banach space?
– ABIM
Dec 17, 2018 at 22:27

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 $$|f(x)-f(y)| \le C d(x,y) \text{ for all } x,y \in X$$ 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 $$L(f_a)\le1$$. Thus $$|f_a|\le1$$. 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 $$\delta_{a_1}, \dots ,\delta_{a_n}$$ are linearly independent. If

$$c_1 \delta_{a_1} + \dots + c_n \delta_{a_n} = 0$$

then consider the function $$g(x) = d(x,\{a_2,...,a_n\})$$. Function $$g$$ is a Lipschitz function. Note that

$$(c_1 \delta_{a_1} + \dots + c_n \delta_{a_n}) (g) = 0,$$ $$c_1 g(a_1) + c_2 g(a_2) + \dots + c_n g(a_n) = 0,$$

$$c_1 g(a_1) = 0$$ which implies that $$c_1=0$$. In a similar fashion one can show that $$c_2, \dots ,c_n$$ must be all 0.

This finishes the proof.

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– mrp
Oct 4, 2016 at 8:56
• Also standard Kuratowski's embedding theorem guarantees that the metric space X embeds into a Banach space, namely CB(X), continuous bounded functions from X to R with sup-norm. We first fix z. Given y, define phi_y from X to R by phi_y(x) = d(x,y) - d(z,x). One can easily verify that phi is an isometry. Also phi_z = 0. But the image of X-{z} in CB(X) is linearly independent. Oct 4, 2016 at 10:40
• Compared to the Kuratowski embedding ($K:X\to \ell^\infty(X)$, $x\mapsto d(x,\cdot)- d(z,\cdot)$) the construction in Ali Kavruk's answer has the virtue of givng a functor from the category of pointed metric spaces (i.e., a metric space with a base point $z$ used to define the norm of L(X)) and base point preserving Lipschitz maps to the category of Banach spaces and continuous linear operators by assigning to a basepoint preserving Lipschitz function $h:X\to Y$ the transposed of the composition opeator $C_h:L(Y)\to L(X)$, $g\mapsto g\circ h$. Nov 7, 2023 at 16:55
• I doubt that also the Kuratowski embedding can be made a functor. Nov 7, 2023 at 16:57

Hint: Perhaps try the continuous real-valued functions on $X$ with a suitable norm.