Finite sets in $\mathbb{R}$ closed under multiplication I have recently started learning Abstract Algebra, starting with binary operations and read about the concept of a set closed under some operation. One of the problems suggested for solving was this one:
Find all finite set in $\mathbb{R}$ such that they are closed under multiplication. I managed to find a solution but it is quite long and sloppy and I will write it in an answer. I would like to see other solutions as I really don't like mine.
 A: Let $A\subset \Bbb{R}$ is a finite set which closed under multiplication.
Claim : $A\subset \{0,\pm1\}$
Assume the contrary that $A\not\subset \{0, \pm 1\}$ . Then $\exists a\in A$ such that $a\neq 0,\pm 1$.
By closure of multiplication $a^n\in A$ and $a^m\neq a^n \quad \forall m\neq n$.
Hence infinite set $\{a^n:n\in\Bbb{N}\}\subset A$ , contradict $A$ is finite.

Number of subsets of $\{0,\pm1\}$ is $2^3=8$.
Now it's easy to find which subsets of $\{0,\pm1\}$ are closed under multiplication.
$\{0\}, \{1\}, \{0, 1\}, \{-1, 1\}, \{0, \pm1\}$
A: Let $a\in A$. Suppose $a\ne 0,1,-1$. If $|a|>1$ then $|a|<|a^2|<|a^3|< \cdots$. If $|a|<1$ then $|a|>|a^2|>|a^3|>\cdots >0$. In any case, $\{a,a^2,a^3, \cdots\}$ is  an infinite set contained in $A$. This contradiction shows that $A\subseteq \{-1,0,1\}$
Edit.
I see you missed one subset. There are $6$ finite subsets of $\Bbb R$ that are closed under multiplication: $\emptyset$, $\{0\}$, $\{1\}$, $\{0,1\}$, $\{-1,1\}$ and $\{-1,0,1\}$
A: Let $A = \{a_1, a_2, \dots, a_n\}$.
If there exist some $a_i > 1$, we can choose the largest one and, since $A$ is closed under multiplication, $a_i^2 \in A$, but $a_i^2>a_i$, thus contradicting that $a_i$ is the maximum number in $A$. Similarly, $a_i \ge -1$, for every $i \in \{1,2, \dots, n\}$.
Then, obviously, $a_1a_2\dots a_n \in A$, so there is a $j \in \{1, 2, ..., n\}$ such that $a_j = a_1a_2 \dots a_n$. If $a_j \neq 0$, $a_1a_2 \dots a_{j-1}a_{j+1} \dots a_n = 1$, so $a_1a_2 \dots a_{i-1}a_{i+1} \dots a_{j-1}a_{j+1} \dots a_n = \frac{1}{a_i} \in A$, for every $i \in \{1, 2, \dots j - 1, j + 1 \dots, n\}$. If $a_j = 0$, we can a similar judgement for $A \setminus \{a_j\}$.
If we suppose $a_i \in (-1, 1) \setminus \{0\}$, $i \in \{1, 2, \dots j - 1, j + 1, \dots, n\}$, then $\frac{1}{a_i} \in (-\infty, 1) \cup (1, \infty)$, which is absurd, since we proved that every elements of A is in $[-1, 1]$. So, $a_i \in \{-1, 0, 1\}$, for every $i \in \{1, 2, \dots, j - 1, j + 1 \dots, n\}$. So, $A = \{-1, 0, 1, a_j\}$. It is easy to see that $a_j \in \{-1, 0, 1\}$ too.
After checking every subset of $\{-1, 0, 1\}$, we arrive at the conclusion that: $$A \in \{\{0\}, \{1\}, \{0, 1\}, \{-1, 1\}, \{-1, 0, 1\}\}$$
I hope my solution is correct, as I am not too sure about the part where I proved $a_i \in \{-1, 0, 1\}$, for every $i \in \{1, 2, \dots, j - 1, j + 1, \dots n\}$
