Is it possible to show hard thresholding of a random vector only produces one vector? The hard thresholding operator $H_s(\mathbf{x})$ keeps the $s$ largest entries of $\mathbf{x} \in \mathbb{R}^n$ in absolute values where $s < n$.
For example for $\mathbf{x}=[1,-3,1]^{\top}$, $H_2([1,-3,1]^{\top})$ is either $[0,-3,1]^{\top}$ or $[1,-3,0]^{\top}$. Therefore, in general, $H_s(\mathbf{x})$ is a set. However, when $\mathbf{x} \in \mathbb{R}^n$ is a continuous random vector, $\mathbb{P}(x_i=x_j)=0$ for all $i \neq j$. This implies $H_s(\mathbf{x})$ is a singleton (only one vector) for any random vector $\mathbf{x}$.
Question 1
Am I right?
Question 2
If so, suppose $\mathbf{x}=[x_1, \dots,x_n]^{\top}$ is a random vector in $\mathbb{R}^n$ where $x_i \sim \mathcal{D}_i$ such that $\mathcal{D}_i$ is an unknow continuous distribution (if it helps we can assume $x_i \sim \mathcal{N}(\mu_i, \sigma_i$) for all $i=1,\dots,n$). Can we show $H_s(\mathbf{x})$ is a singleton (only one vector)?
Note
This question asks whether we can order elements of a continuous random vector or not. Or, whether elements of a this random vector are always not equal.
 A: Just to clear up the terminology a bit.  $H_s(x)$ is a set, representing all possible results of the thresholding operation (for that specific $x$).  $|H_s(x)|$ is the size of that set, and $|H_s(x)| = 1$ means the set is a singleton.
Claim: If $\forall i \neq j: P(x_i = x_j) = 0$, then $P(|H_s(x)| = 1) = 1$.
Proof:
$$
\begin{array}{ccc}
|H_s(x)| > 1 &\implies &\exists i \neq j: x_i = x_j 
\\
P(|H_s(x)| > 1) &\le &P(\exists i \neq j: x_i = x_j)
\\
 &= &P(\bigcup_{i \neq j} x_i = x_j)
\\
 &\le &\sum_{i \neq j} P(x_i = x_j) &\text{by union bound}
\\
 &= &0
\end{array}$$

So if your distributions $D_i$ meet the precondition that $\forall i \neq j: P(x_i = x_j) = 0$, then $H_s(x)$ is a singleton with probability $1$.
In particular, one way to meet the precondition is this: if all $D_i$ are independent, and if every $D_i$ has a CDF with no steps, i.e. its pdf has no "impulse", i.e. $\forall i, \forall v \in \mathbb{R}, P(x_i = v) = 0$.  The independence and the lack of impulse together meet the precondition, and so the claim holds and you can conclude $P(|H_s(x)|=1) = 1$.
Does this answer your question, or am I answering the wrong question? :)
