From the set theory perspective, let's note
$$P \overset{\text{def}}{=}\{p \in \mathbb{N} \mid p \text{ - prime}\}$$
$$P_{\leq n} \overset{\text{def}}{=} \{p \in P \mid p \leq n \}$$
$$M_n \overset{\text{def}}{=} \{p \in P \mid p \mid n\}$$
$\color{red}{M_n \ne \varnothing, \forall n\geq2}$. Then, it's easy to show
$$P_n=P_{\leq n} \setminus M_n \tag{1}$$
$$M_n \subset P_{\leq n} \tag{2}$$
Now, let's assume $\exists n \ne m$, in fact we can assume $\color{red}{n>m}$, such that $\color{blue}{P_n = P_m}$ then
$$P_{\leq n} \setminus M_n = P_{\leq m} \setminus M_m \overset{(2)}{\Rightarrow} P_{\leq n}=\left(P_{\leq n} \setminus M_n \right) \bigcup M_n=\left(P_{\leq m} \setminus M_m \right) \bigcup M_n$$
or
$$P_{\leq n}=\left(P_{\leq m} \setminus M_m \right) \bigcup M_n \tag{3}$$
given $\color{red}{P_{\leq m} \subset P_{\leq n}}$ and $(2)$, it's easy to deduct that
$$M_m \subset M_n \tag{4}$$

There are 3 possible cases to exploit with $(4)$:

**Case 1 (incomplete).** $\forall k \in M_n, k\in M_m$ or $M_n=M_m$ (like $n=2^2 \cdot 3^2, m=2 \cdot 3$ for example), then $\color{green}{P_{\leq n} = P_{\leq m}}$. If $m=\prod\limits_{i} p_{k_i}^{\alpha_i}$ then $n=\prod\limits_{i} p_{k_i}^{\beta_i}$ with $\beta_i \geq \alpha_i$ and at least one $\beta_{i^{*}} > \alpha_{i^{*}}$. Since all $p_i \geq 2$ this means $$m < 2m \leq n$$ From Bertrand postulate there is $p^{*}$-prime in between $m$ and $2m$ with $p^{*} \notin P_{\leq m}$ and $p^{*} \in P_{\leq n}$, thus $\color{green}{P_{\leq m} \ne P_{\leq n}}$ - **contradiction**.

*Obviously $\beta_i \geq \alpha_i$ doesn't cover all the cases, e.g. $n=2 \cdot 3^2, m=2^2 \cdot 3$.*

**Case 2.** $\exists k \in M_n, k\notin M_m$ and $k \in P_{\leq m}$. Since $k \in P_{\leq m}$ and $k\notin M_m$ then $k \in P_m=P_{\leq m} \setminus M_m$, but because $k \in M_n$ and $k \in P_{\leq n}$ then $k \notin P_n=P_{\leq n} \setminus M_n$. Thus $\color{blue}{P_n \ne P_m}$ - **contradiction**.

**Case 3.** $\exists k \in M_n, k\notin M_m$ and $k \notin P_{\leq m}$. Then $n=k\cdot t \geq k > m$

- if $t=1$ then (given $k$ is prime) $n$ is prime and $P_n$ will contain all the prime factors of $m$, but $P_m$ will not contain prime factors of $m$, thus $\color{blue}{P_n \ne P_m}$ -
**contradiction**.
- if $t\geq2$ then between $n$ and $\frac{n}{2}$ exist a prime $p^{*}$ (
**proposition 1**, below). Because $p^{*}>\frac{n}{2}=\frac{kt}{2}
\geq k>m \Rightarrow p^{*} \notin P_m$. Also $p^{*} \nmid n$, otherwise $n=qp^{*}\geq 2p^{*} > n$. Thus $p^{*} \in P_n$ and as a result $\color{blue}{P_n \ne P_m}$ - **contradiction**.

**Proposition 1.** $\forall n \geq 3$ $\exists p$ - prime s.t. $\frac{n}{2}<p<n$

Let's take $k=\left \lfloor \frac{n}{2} \right \rfloor+1 \geq \frac{n}{2}$. Bertrand postulate says $\exists p$ - prime s.t. $k < p < 2k-2$ or
$$\color{red}{\frac{n}{2}\leq k} < p < 2k-2=\color{blue}{2 \left \lfloor \frac{n}{2} \right \rfloor \leq n} \tag{5}$$
simply because:

- for $n=2r$ (even) we have $\color{red}{\frac{n}{2}=r < r+1=\left \lfloor \frac{n}{2} \right \rfloor+1=k}$ and $\color{blue}{2 \left \lfloor \frac{n}{2} \right \rfloor=2r=n}$
- for $n=2r+1$ (odd) we have $\color{red}{\frac{n}{2}=r+\frac{1}{2}<r+1=\left \lfloor r+\frac{1}{2} \right \rfloor+1=\left \lfloor \frac{n}{2} \right \rfloor+1=k}$ and $\color{blue}{2 \left \lfloor \frac{n}{2} \right \rfloor=2 \left \lfloor r+\frac{1}{2} \right \rfloor=2r<2r+1=n}$

Bertrand postulate says $(5)$ is true for $k>3$ or $n >4$. For $n=4$ we have $\frac{4}{2}<3<4$ and for $n=3$ we have $\frac{3}{2}<2<3$.