How to prove if $A \times C \subseteq B \times D \implies A \subseteq B$

Question

My proof

Given $(x,y) \in A \times C \implies x \in A$ and $y \in C$ since $A \times C \subseteq B \times D$ then $(x,y) \in B \times D$ then $x \in B$ and $y \in D$

since $x \in A$ and $x \in B$ then $A \subseteq B$

Is this a appropriate proof?

Sorry typo in title has been edited.

Answer 1

To show that $A \subseteq B$, you need to start by choosing any $x \in A$ and conclude by showing that $x \in B$. Notice that the implication is false if $C = \varnothing$; indeed, $\{1\} \times \varnothing = \varnothing \subseteq \{7\} \times \{7\}$ but $\{1\} \not\subseteq \{7\}$.

Otherwise, choose any $x \in A$ and suppose that there is some $y \in C$. Then $(x, y) \in A \times C$. But since $A \times C \subseteq B \times D$, we know that $(x, y) \in B \times D$ so that $x \in B$, as desired.

Answer 2

You should check your question again since the inference you have written is false.

Let $A = \mathbb{Q}$, $B = \mathbb{R}$, $C = \mathbb{N}$, and $D = \mathbb{N}$.

Then $A \times C \subseteq B \times D$ however $\mathbb{Q} \not \subseteq \mathbb{N}$.