Assume that we have a vector $x$ with dimension $3$. The value of $x = {4, 2, 8}$. My question is if there is a norm that can generate one value of $x$ that are unique compared to other samples of $x$.
For example. Using $L^2$ norm generates the same number if we have two $x$ e.g $x_1 = {-4, 2, 8}$ and $x_2 = {4, 2, 8}$.
$$||x_1||_2 = ||x_2||_2$$
And if I'm using another values of $x$ e.g $x_3 = {8, 2, 4}$ then $L^1$ norm would be the same for $$||x_2||_1 = ||x_3||_1$$
So is there a norm that can create a very unique number of a random vector $x$?
Here's one way to construct a one-one map from the set of all ordered triples of reals to the reals:
Write your triple of reals as $(a+r,b+s,c+t)$ with $a,b,c$ integers and $r,s,t$ between zero (inclusive) and one (exclusive). Map it to the real whose integer part is $2^{|a|}3^{|b|}5^{|c|}7^d11^e13^f$ where $d$ (respectively, $e$, $f$) is zero if $a$ (respectively $b$, $c$) is at least zero; otherwise, one; and the decimal part is obtained by interleaving the digits of $r,s,t$. That is, if $r=.r_1r_2\dots$, $s=.s_1s_2\dots$, $t=.t_1t_2\dots$, then you form $.r_1s_1t_1r_2s_2t_2\dots$.
For example, given the triple $(\pi,-e,-sqrt2)=(3.14159\dots,-2.71828\dots,-1.41421\dots)$, rewrite it as $(3+.14159\dots,-3+.28171\dots,-2+58578)$, and map it to $2^33^35^2\times11\times13+.125488115577918\dots=772200.125488115577918\dots$.
you could do the same thing in binary, or any base you like.