# If $S\subseteq\Bbb R$ is closed under multiplication, dividing $S$ into two parts with some property, one of them is closed under multiplication

Let $$S$$ be a subset of $$\mathbb{R}$$ that is closed under multiplication (that is, if $$a$$ and $$b$$ are in $$S$$, then so is $$ab$$). Let $$T$$ and $$U$$ be disjoint subsets of $$S$$ whose union is $$S$$. Given that the product of any three (not necessarily distinct) elements of $$T$$ is in $$T$$ and that the product of any three elements of $$U$$ is in $$U$$, show that at least one of the two subsets $$T$$, $$U$$ is closed under multiplication.

So I'm trying to do a proof by contradiction, except that I'm not yet very familiar with it, and I'm not sure if my proof (below) is right... please tell me what I can fix.

We start out with the assumption that neither $$T$$ nor $$U$$ is closed under multiplication and use proof by contradiction to prove that at least one of $$T$$ or $$U$$ must be closed under multiplication.

If $$t_1, t_2, t_3 \in T$$ and $$u_1, u_2, u_3 \in U$$, it is given by the problem that $$t_1 \cdot t_2 \cdot t_3 \in T$$ and $$u_1 \cdot u_2 \cdot u_3 \in U$$. Also, $$T$$ and $$U$$ are disjoint subsets of $$S$$ and therefore, $$T \cap U$$ = $$\varnothing$$.

Now let $$t_3 = u_1 \cdot u_2$$ and $$u_3 = t_1 \cdot t_2$$. These statements are valid because $$T$$ and $$U$$ are not closed under multiplication, so the product of $$u_1 \cdot u_2$$ must not be in the set $$U$$ and the product of $$t_1 \cdot t_2$$ must not be in set $$T$$.

(actually, I'm confused here because I feel "the product of $$t_1 \cdot t_2$$ must not be in set $$T$$" is already a contradiction as it is given that "the product of any three (not necessarily distinct) elements of T is in T" and the elements could be all 1.)

Then $$t_1 \cdot t_2 \cdot t_3 \in T$$ is equivalent to $$t_1 \cdot t_2 \cdot u_1 \cdot u_2 \in T$$ and $$u_1 \cdot u_2 \cdot u_3 \in U$$ is equivalent to $$u_1 \cdot u_2 \cdot t_1 \cdot t_2 \in U$$. However, this is a contradiction because these products both are the same —$$t_1 \cdot t_2 \cdot u_1 \cdot u_2$$— but $$T \cap U$$ = $$\varnothing$$ because $$T$$ and $$U$$ are disjoint. Then the original assumption that "neither $$T$$ nor $$U$$ is closed under multiplication" must have been wrong, and at least one of $$T$$ or $$U$$ must be closed under multiplication.

• "$T$ is closed under multiplication" reads "$ab \in T$ for all $a,b \in T$". Its negation, "$T$ is not closed under multiplication", reads "there is $a,b \in T$ such that $ab \notin T$", not "$ab \notin T$ for all $a,b \in T$". Dec 27, 2017 at 4:14
• That is not already a contradiction because $1$ may not be in $T$. Dec 27, 2017 at 4:15