# Need help putting proof into proper form

• Jun 29th 2008, 04:35 PM
Ultros88
Need help putting proof into proper form
Show that the relations $(A \cup C) \subset (A \cup B)$ and $(A \cap C) \subset (A \cap B)$, when considered together, imply $C \subset B$.

Here is my proof:
If $x \in (A \cup C)$ then $x \in (A \cup B)$.
If $x \in (A \cup C)$ then $x \in A$ or $x \in C$.
We want to know about the elements in C so we consider $x \in C$.
If $x \in C$ then $x \in (A \cup B)$.
This means that if $x \in C$ then $x \in A$, $x \in B$, or $x \in A$ and $x \in B$.
There are three cases.
First: $x \in C, x \in A \implies x \in (A \cap C)$. From the second relation, $x \in (A \cap B)$, and therefore $x \in B$.
Second: $x \in C, x \in B \implies x \in B$.
Third: $x \in C, x \in (A \cap B) \implies x \in B$.
Thus $x \in C \implies x \in B$, therefore $C \subset B$.

Is this proof complete, correct, clean, nasty?
Any help is appreciated, especially cleaning it up and making it look all nice with proper notation. Again, if it's wrong it would be more helpful if you could point out where I've went astray.
Thanks,
ultros
• Jun 29th 2008, 06:38 PM
Ultros88
Disproving Converse
Also, I'm having difficulty finding an example to disprove the converse. Thanks again.
• Jun 29th 2008, 08:05 PM
Isomorphism
Quote:

Originally Posted by Ultros88
Also, I'm having difficulty finding an example to disprove the converse. Thanks again.

I think the converse is true:

That is:

If $C \subset B, A \cup C \subset A \cup B$ and $A \cap C \subset A \cap B$

I think its easy to prove:

$x \in A \cup C \Rightarrow x \in A \text{ or } x \in C \subset B \Rightarrow x \in A \text{ or } x \in B \Rightarrow x \in A \cup B$

Similarly the other part.

P.S: What inspired the name Ultros(FF?)?
• Jun 30th 2008, 11:49 AM
Ultros88
Converse
Yeah, I think the converse is true as well, which is funny because the book I'm using seems to want me to prove it false.

And the name did come from FF by the way. My favourite octopus.