1. $A\subseteq C, B\subseteq D \Rightarrow A \times B \subseteq C \times D$

  2. $A,B\neq \emptyset \Rightarrow A\times B \subseteq C\times D \Rightarrow A\subseteq C, B\subseteq D$

  1. $\forall x\in A : \exists x\in C \\ \forall y\in B : \exists y\in D \\ \Rightarrow \forall(x,y)\in A\times B :\exists (x,y)\in C\times D \\ \Rightarrow A\times B \subseteq C\times D$

But with 2. if it weren't for $A,B\neq \emptyset$ it would be false but now I think it's basically the same as my proof for 1. only in the opposite way. Any thoughts please ?


The existential quantors mentioned in your question must be left out.

1) Let $(x,y)\in A\times B$ or equivalently $x\in A$ and $y\in B$. Then $x\in C$ as a consequence of $A\subseteq C$ and $y\in D$ as a consequence of $B\subseteq D$. That means that $(x,y)\in C\times D$. This proves that $A\times B\subseteq C\times D$.

2) Let $B\ne\emptyset$ and let $x\in A$. Some $y\in B$ exists and automatically $(x,y)\in A\times B\subseteq C\times D$ so that $x\in C$. This proves that $A\subseteq C$ and likewise it can be proved that $A\ne\emptyset$ leads to $B\subseteq D$.

If $B=\emptyset$ then automatically $A\times B=\emptyset\subseteq C\times D$. It is not necessary here that $A\subseteq C$.

Analogous story if $A=\emptyset$.

  • $\begingroup$ You are very welcome. $\endgroup$ – drhab Apr 27 '14 at 15:39

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