Let us define the following :
the pseudo-power set $P = \{A \subseteq \mathbb{Z} \text{ : }A \text{ is a finite set}\}$
the function $d : P \times P \to \mathbb{R}$ as $d(A,B) = \text{cardinality of } (A\cup B)\setminus (A\cap B) = A \Delta B$
Then it is not difficult to observe that $(P,d)$ forms a metric space.
My question is : Is it possible to define an isometry from $(P,d)$ into $\mathbb{R}^n$ for any $n \in \mathbb{N}$ ?
I tried but couldn't come up with any such isometry and hence I believe the answer must be NO. But I would love a formal rigorous proof for this. Can anyone please help ?
An easy topological argument: Note that $d(\emptyset, \{a\})=1$, hence $\{\{a\}: a\in\mathbb Z\}$ is bounded. If it embeds into $\mathbb R^n$, then it has a Cauchy subsequence $v_{n_i}$, thus $d(v_{n_i}, v_{n_j})<1$ for sufficiently large $i, j$ while $d(\{n_i\}, \{n_j\}) = 2$ if $n_i\not=n_j$.
WLOG, we may assume $\emptyset$ is mapped to $0$, and for any $\{a\}$, we denote its image by $v_a$. Then for any pair of distinct singlestons $\{a\}, \{b\}$, the triangle formed by $0, v_a, v_b$ must be degenerate as the triangle inequality $d(\{a\}, \emptyset) + d (\{b\}, \emptyset) = d(\{a\}, \{b\})$ is tight. Therefore the $v_a$ and $v_b$ must be linearly dependent, so $v_a = cv_b$, $1 = \|v_a\| = |c| \|v_b\| = |c|$, but $v_a\not=v_b$, so $c=-1$. Therefore $v_1 = -v_0$ and $v_2=-v_0$, so $v_1=v_2$, a contradiction. (More concisely, we can employ the parallelogram identity $2\|u\|^2 + 2\|v\|^2 = \|u-v\|^2 + \|u+v\|^2$ to conclude $u+v=0$ in our case.)
This shows that $\emptyset, \{0\}, \{1\}, \{2\}$ cannot be embedded into any $\mathbb R^n$. (In fact, with a little bit more effect, we can show this finite metric space cannot be isometrically embedded into any Hilbert space $\mathcal H$ -- real or complex, finite or infinite dimensional.)
Here is a linear algebra argument which is actually the first that comes to my mind.
By taking singletons, $P$ has infinitely many points $a_1 = \{1\}, a_2 = \{2\}, \cdots$ such that $d(a_i, a_j)=2\delta_{ij}$ for any $i, j$. By rescaling, we may ignore the factor $2$, and assume $d(a_i, a_j)=\delta_{ij}$. For further convenience, we may assume $a_1$ is mapped to $0\in\mathbb R^n$ in a potential embedding. Now we show that $b_0=0, b_1, \cdots, b_m$ where $d(b_i, b_j)=\delta_{ij}$ can only exist for $m\le n$, therefore no $\mathbb R^n$ is good enough for the embedding.
To show $m\le n$, we show that $\{b_i\}_{i=1}^m$ are linearly independent. Indeed, by $d(b_0, b_i)=1\Leftrightarrow \|b_i\|=1$ and $\sqrt{(b_i-b_j, b_i-b_j)}=d(b_i, b_j)=\delta_{ij}$, we can compute the inner products $(b_i, b_j) = 1/2$ for $0<i<j$.
Therefore the matrix $(b_{ij})(b_{ij})^t = \begin{pmatrix} 1 & 1/2 & \cdots \\ 1/2 & 1 & \cdots \\ \vdots & \vdots & \vdots \end{pmatrix} = \frac{1}{2}(I+A)$, where $A$ is the matrix with all entries being $1$. $A$ only has eigenvalues $0$ and $m$, thus $(b_{ij})(b_{ij})^t$ only has eigenvalues $\frac{1}{2}$ and $\frac{1+m}{2}$, so it's invertible and $b_1, \cdots, b_m$ must be linearly independent.