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?

  • $\begingroup$ The set you constructed is NOT a basis for $\Bbb R^X$. $\endgroup$ – PVAL-inactive Oct 29 '14 at 18:52
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    $\begingroup$ 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 $d$. $\endgroup$ – polmath Oct 29 '14 at 18:54
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    $\begingroup$ Probably useful: en.wikipedia.org/wiki/Kuratowski_embedding $\endgroup$ – Nate Eldredge Oct 29 '14 at 18:57
  • $\begingroup$ 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? $\endgroup$ – AIM_BLB Dec 17 '18 at 22:27

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


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 R, denoted by L(X). Recall that a function f:X-->R is called Lipschitz if there exists a positive constant C such that

|f(x)-f(y)| =< C d(x,y) for all x,y.

The infimum of all such constants is denoted by L(f). L(f) is not a norm on L(X) (for example 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 |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{|F(f)|: f is in L(X) with |f(z)|+L(f) =< 1 }.

We first observe that point evaluations are continuous: if a is in X then delta_{a} given by delta_{a} (f) = f(a) has norm less than or equal d(z,a)+1. In fact

|delta_{a}| = sup{|f(a)|: |f(z)|+L(f) =< 1} =< sup{|f(a)-f(z)|+|f(z)|: ........} =< sup{L(f)d(a,z) + |f(z)|: ....... } =< d(a,z)+1 (note: so both |f(z)| and L(f) must be =< 1).

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: Let's fix a and b. First note that

|delta_a-\delta_b| = sup{|f(a)-f(b)|: |f(z)| + L(f) =< 1 } =< sup{|f(a)-f(b)|: L(f) =< 1} =< d(a,b).

Now, consider the function f(x) = d(x,a)-d(a,z). Then f(z)=0 and one can easily shows that L(f) =< 1. Thus |f| =< 1. Moreover it is easy to show that |f(a)-f(b)| = d(a,b). This shows that |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. In fact 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.

  • $\begingroup$ Welcome to MSE! Please see this tutorial on how to typeset your posts so they are easier to read. $\endgroup$ – mrp Oct 4 '16 at 8:56
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    $\begingroup$ 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. $\endgroup$ – Ali Kavruk Oct 4 '16 at 10:40

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