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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.

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  • $\begingroup$ "$T$ 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 $a,b \in T$". $\endgroup$
    – Kenny Lau
    Dec 27, 2017 at 4:14
  • $\begingroup$ That is not already a contradiction because $1$ may not be in $T$. $\endgroup$
    – Kenny Lau
    Dec 27, 2017 at 4:15
  • $\begingroup$ In symbols, my first comment reads: the negation of "$\forall a,b \in T: ab \in T$" is "$\exists a,b \in T: ab \notin T", not "$\forall a,b \in T: ab \notin T$" $\endgroup$
    – Kenny Lau
    Dec 27, 2017 at 4:26
  • $\begingroup$ At the point where you say you are confused " .... because the elements could all be $1$." you should note that there is no reason why $1$ must belong to $S$. $\endgroup$ Dec 27, 2017 at 7:50

2 Answers 2

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Your proof is almost correct, in the sense that the outline is correct, but there are some logical mistakes in between.

If $U$ is not closed under multiplication, it does not mean "for every $u_1, u_2 \in U$, $u_1 u_2 \notin U$", but rather, "there exists $u_1, u_2 \in U$ such that $u_1 u_2 \notin U$.


So, here is my take on your proof, with copied parts in quotes (yellow box) and my redactions outside:


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.

Since $T$ is not closed under multiplication, there are $t_1$ and $t_2$ both in $T$ such that their product, $t_1 \cdot t_2$, is not in $T$. Since $T$ and $U$ are disjoint and span $S$ which is closed under multiplication, $t_1 \cdot t_2$ must be in $U$.

Similarly, there are $u_1$ and $u_2$ both in $U$ with the property that $u_1 \cdot u_2 \in T$.

Now let $t_3 = u_1 \cdot u_2$ and $u_3 = t_1 \cdot t_2$.

Since $t_1$, $t_2$, and $t_3$ are all in $T$, by the assumption that $T$ is closed under triple product, we have $t_1 \cdot t_2 \cdot t_3 \in T$. Similarly, $u_1 \cdot u_2 \cdot u_3 \in U$.

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.

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  • $\begingroup$ Regarding the part where you said "If U is not closed under multiplication, it does not mean...",does that mean that that for other values of $u_1$ and $u_2$, $u_1u_2$ could be in U? $\endgroup$
    – space
    Dec 27, 2017 at 8:14
  • $\begingroup$ @Helena that is correct. $\endgroup$
    – Kenny Lau
    Dec 27, 2017 at 8:15
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SAMPLE PROOF. For comparison.

Suppose by contradiction that neither $T$ nor $U$ is closed under multiplication.

Let $t_1,t_2\in T$ with $t_1t_2\not \in T.$ Then $t_1t_2\in S$ \ $T=U.$ Let $t_1t_2=u_3.$

Let $u_1,u_2\in U$ with $u_1u_2\not \in U.$ Then $u_1u_2\in S$ \ $U=T.$ Let $u_1u_2=t_3.$

Then $u_1 u_2 t_1 t_2=u_1 u_2 u_3 \in U$ because $u_1,u_2,u_3\in U$,

and $t_1 t_2 u_1 u_2=t_1 t_2 t_3 \in T$ because $t_1,t_2,t_3\in T$.

So $u_1u_2t_1t_2=t_1t_2u_1u_2\in U\cap T,$ contradicting $U\cap T=\phi.$

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