Answer: Three-Fold Closure Forces Two-Fold Closure

Key Idea / Intuition

Suppose both $T$ and $U$ fail to be closed. Then some pair from $T$ "leaks" into $U$, and some pair from $U$ leaks into $T$. Multiply both leaky pairs together to get four numbers. The same product can be bracketed as "one element of $U$ times two elements of $U$" or as "two elements of $T$ times one element of $T$." The three-fold rule then forces this single number to lie in both $U$ and $T$, which is impossible because they are disjoint.

The whole proof is: one product, two bracketings, two incompatible verdicts.

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Formal Proof / Solution

Step 1: Assume both fail. Suppose, for contradiction, that neither $T$ nor $U$ is closed under multiplication. Then:

• there exist $t_1,t_2\in T$ with $t_1t_2\notin T$;
• there exist $u_1,u_2\in U$ with $u_1u_2\notin U$.

Step 2: Locate the leaked products. Since $S$ is closed under multiplication, $t_1t_2\in S$ and $u_1u_2\in S$. Since $S=T\sqcup U$, it follows that $t_1t_2\in U,\qquad u_1u_2\in T.$

Step 3: Bracket the same product two ways. Consider $P = t_1t_2u_1u_2 .$

• First bracketing. Write $P=(t_1t_2)\cdot u_1\cdot u_2$. Each of the three factors $t_1t_2,\;u_1,\;u_2$ lies in $U$. By the three-fold hypothesis for $U$, we get $P\in U$.
• Second bracketing. Write $P=t_1\cdot t_2\cdot(u_1u_2)$. Each of the three factors $t_1,\;t_2,\;u_1u_2$ lies in $T$. By the three-fold hypothesis for $T$, we get $P\in T$.

Both bracketings are legitimate because multiplication of real numbers is associative.

Step 4: Contradiction. We have found $P\in T\cap U$, but $T$ and $U$ are disjoint. Hence our assumption was false, and at least one of $T$, $U$ is closed under multiplication. $\blacksquare$

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Remarks

• What was actually used. We used only that multiplication is associative. Commutativity was not needed, and neither was the fact that we are working with real numbers. The same argument works in any semigroup $S=T\sqcup U$.
• The model example. Positives versus negatives is the prototype. There, $U$ behaves like the "odd" coset of a $\mathbb{Z}/2$-grading. Three-fold closure is exactly what an odd coset satisfies, since odd + odd + odd is odd. The theorem says that two such "odd-like" pieces cannot coexist.
• The general lesson. When a hypothesis is about $k$-fold products and the conclusion is about $2$-fold products, try building a single product that can be read as a $k$-fold product in two different ways.