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The problem is: prove if A, B, and C are sets and $A \times C = B \times C$, then A = B.

I decided to prove it and wrote this proof:

Proof. First we will show that if A, B, and C are sets and $A \times C = B \times C$, then $A \subseteq B$. Assume $(x,y) \in A \times C$. Thus, $x \in A$ by definition of cartesian product. Hence, $(x,y) \in B \times C$ by definition of subset (equal sets are subsets of each other). Consequently, $x \in B$ by definition of cartesian product. Therefore $x \in A$ implies $x \in B$, so it follows that $A \subseteq B$.

Conversely, we will show that if A, B, and C are sets and $A \times C = B \times C$, then $B \subseteq A$. Assume $(x,y) \in B \times C$. Thus $x \in B$ by definition of cartesian product. Hence $(x,y) \in A \times B$ by definition of subset (equal sets are subsets of each other). Consequently, $x \in A$ by definition of cartesian product. Therefore, $x \in B$ implies $x \in A$, so it follows that $B \subseteq A$.

Hence, we have shown that $A \subseteq B$ and $B \subseteq A$, so $A = B$.

IS IT CORRECT?

mont2223
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    ${0}\times\varnothing={0,1}\times\varnothing$, but ${0}\ne{0,1}$. – Brian M. Scott Apr 15 '21 at 01:08
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    The error in your proof lies with the line "Assume $(x,y)\in A\times C$" This assumption might not have been valid in the case that $A\times C$ were empty. If you were to require $A,B,C$ all to be non-empty then your proof would have at a glance been fine. – JMoravitz Apr 15 '21 at 01:12
  • You need $C$ not empty for this to be true. – Thomas Andrews Apr 15 '21 at 02:18

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