# A Proof in Elementary Set Theory

I apologize in advance for any lack of clarity in my mathematical symbols; I'm beginning to learn how to use this site.

I'm working through "Introduction to Analysis" by Rosenlicht and he presents an exercise

Prove that if $X\subset S$, $Y\subset S$, then $\complement X\cap\complement Y=\complement(X\cup Y)$

I went about it as follows:

Suppose $X\subset S$ and $Y\subset S$.   Then $\complement X=\{x:x\in S\land x\notin X\}$ And $\complement Y=\{x:x\in S\land x\notin Y\}$ , so $\complement X\cap\complement Y=\{x:x\in S\land x\notin X\land x\in S \land x\notin Y\}$.   Hence $\complement X\cap\complement Y=\{x:x\in S\land x\notin X\land x\in Y\}$.   Therefore $\complement X\cap\complement Y=\{x:x\in S\land x\notin X\cup Y\}$  And thus $\complement X\cap\complement Y=\complement(X\cup Y)$

Whereas the author gives the proof

If $x\in \complement X\cap\complement Y$ then $x\in\complement X$ and $x\in \complement Y$.   This means that $x\in S, x\notin X, x\notin Y$.   Since $x\notin X, x\notin Y$, we know that $x\notin X\cup Y$.   Hence $x\in \complement(X\cup Y)$.

Conversely, if $x\in\complement(X\cup Y)$, then $x\in S$ and $x\notin X\cup Y$.   Therefore $x\notin X$ and $x\notin Y$.   Thus $x\in\complement X$ and $x\in\complement Y$, so that $x\in\complement X\cap\complement Y$.

My questions are

• 1) is my proof correct?
• 2) which proof is better
• 3) what are the subtle (because obviously I know what the difference is, but I'm looking for deeper mathematical understanding) differences between my proof and the authors?

Thank you!

• If you could please fix your post, in order to be more readable. It's $\complement$ rather than $\compliment$. After each command, you'd better leave a space. If you want to write "x in (not in) Y", you can write either $x \in Y$ or $x \notin Y$. – thanasissdr Aug 20 '15 at 23:30
• Thanks you. That's good to know. It appears someone just did now. Im assuming you can read it now? – Liam Cooney Aug 20 '15 at 23:38
• I think that both proofs are correct (barring a typo at the end of the line following "Hence" in your proof). – Akiva Weinberger Aug 20 '15 at 23:40
• It still needs work. You should also note that logical and is \land – Graham Kemp Aug 20 '15 at 23:41

One additional thing the author does is explicitly show that the implication works both ways; that $x\in(\complement X\cap\complement Y)\implies x\in \complement(X\cap Y)$ and that $x\in\complement(X\cap Y)\implies x\in(\complement X\cap\complement Y)$.   You rely on each step being an equivalence; but don't explicitly indicate this is so.