# [SOLVED] set proof help

• Mar 5th 2009, 10:30 AM
horan
[SOLVED] set proof help
Can anyone prove or disprove
$\displaystyle X \cap Z \subseteq Y \Leftrightarrow (X - Y) \cup (Y-Z) \subseteq Z^c$

I think you have to split it up into two parts, but whenever I do it one way I can't get the same exact argument to go back the other way.
• Mar 5th 2009, 10:52 AM
running-gag
Quote:

Originally Posted by horan
Can anyone prove or disprove
$\displaystyle X \cap Z \subseteq Y \Leftrightarrow (X - Y) \cup (Y-Z) \subseteq Z^c$

I think you have to split it up into two parts, but whenever I do it one way I can't get the same exact argument to go back the other way.

Hi

Suppose that $\displaystyle X \cap Z \subseteq Y$
Let's prove that $\displaystyle (X - Y) \cup (Y-Z) \subseteq Z^c$

Let $\displaystyle x \in (X - Y) \cup (Y-Z)$
We have to prove that $\displaystyle x \in Z^c$

$\displaystyle x \in (X - Y) \cup (Y-Z)$ means that or $\displaystyle x \in (Y-Z)$ or $\displaystyle x \in (X - Y)$
If $\displaystyle x \in (Y-Z)$ then $\displaystyle x \notin Z$ and therefore $\displaystyle x \in Z^c$ => OK
If $\displaystyle x \in (X - Y)$ then $\displaystyle x \in X$ and $\displaystyle x \notin Y$. If $\displaystyle x \in Z$ then $\displaystyle x \in X \cap Z$. But $\displaystyle X \cap Z \subseteq Y$ therefore $\displaystyle x \in Y$ which is not possible since $\displaystyle x \notin Y$. Threfeore $\displaystyle x \notin Z$.
In both cases $\displaystyle x \in Z^c$.
Therefore $\displaystyle (X - Y) \cup (Y-Z) \subseteq Z^c$.
• Mar 6th 2009, 02:41 AM
Set Theory Proof
Hello horan
Quote:

Originally Posted by horan
Can anyone prove or disprove
$\displaystyle X \cap Z \subseteq Y \Leftrightarrow (X - Y) \cup (Y-Z) \subseteq Z^c$

I think you have to split it up into two parts, but whenever I do it one way I can't get the same exact argument to go back the other way.

running-gag has proved that $\displaystyle X \cap Z \subseteq Y \Rightarrow (X - Y) \cup (Y-Z) \subseteq Z^c$. Here's the proof of $\displaystyle X \cap Z \subseteq Y \Leftarrow (X - Y) \cup (Y-Z) \subseteq Z^c$.

Suppose that $\displaystyle (X - Y) \cup (Y-Z) \subseteq Z^c$and that $\displaystyle x \in X\cap Z$. Therefore we must show that $\displaystyle x \in Y$.

$\displaystyle x \in X \cap Z \Rightarrow x \in Z$

$\displaystyle \Rightarrow x \notin Z^c$

$\displaystyle \Rightarrow x \notin (X - Y) \cup (Y-Z)$

$\displaystyle \Rightarrow x \notin (X - Y)$

But $\displaystyle x \in X$

Therefore $\displaystyle x \in Y$

• Jul 17th 2009, 07:43 AM
doresa
hello running-gag
how can you say like that..
• Jul 17th 2009, 10:01 AM
Hello doresa
Quote:

Originally Posted by doresa

You're just quoting part of the whole proof here. Obviously you can't say that, in general, $\displaystyle x \in Z \Rightarrow x \in X \cap Z$.

But the bit you're quoting is from a line that starts: If $\displaystyle x \in (X-Y)$... In other words, it pre-supposes that $\displaystyle x \in X$. Therefore if in addition $\displaystyle x \in Z$, then $\displaystyle x \in X \cap Z$.

Is that OK now?