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1Understand what proof? – Git Gud Apr 07 '14 at 22:49
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3This is not just easy but in fact immediate from the definitions. Why don't you start by writing them down? – Andrés E. Caicedo Apr 07 '14 at 22:50
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Well, $x-Y$ consists of all elements of $X$ that are not in $Y$. Which of these are also in $Y$? – Dirk Apr 07 '14 at 23:14
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Recall the definitions.
- $X\cap Y$ is the set of all $a$ such that $a\in X$ and $a\in Y$.
- $X - Y$ is the set of all $a$ such that $a\in X$ and $a\notin Y$. (Also denoted by $X\setminus Y$ sometimes.)
- $X=Y$ if for every $a$ we have that if $a\in X$, then $a\in Y$ and if $a\in Y$ then $a\in X$. This translates to the statement that $X\subseteq Y$ and $Y\subseteq X$.
So we have to show equality between two sets. One of them is empty, and the empty set is always a subset of every set, so it suffices to show that there are no elements to the set in the left hand side of the equation.
So you have to prove that given any element $a$, it is impossible that $a$ is in $Y$ and in $X-Y$. How should you do that? Unfold the definition. If it is in both sets, then it satisfies both requirements. Can you see why this is impossible?
Asaf Karagila
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I'm not quite sure what you mean there. Yes, these laws are valid, and it seems like a reasonable proof if you approach it from a Boolean algebraic point of view, rather than a set theoretic point of view. (Also note that this is exactly what one of the answers in this page states.) – Asaf Karagila Apr 07 '14 at 23:51
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\begin{align*} (X\setminus Y)\cap Y &= (X\cap Y^C)\cap Y \\ &= X\cap(Y^C\cap Y) \\ &= X\cap\varnothing \\ &= \varnothing \end{align*}
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This is succinct and as simple as possible, completely axiomatic. Much more understandable, and brief, than most of the other explanations I have seen. +1. – MPW Apr 07 '14 at 23:40
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@BrennaRich: Not sure if it has a standard name. Wikipedia calls it "Annihilator for $\cap$". – Apr 07 '14 at 23:47
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This is a pretty trivial proof to understand. If you take anything away from something, and then ask for all the "somethings" back, you will get nothing. This is what the intersection operator is doing, it's requesting all the elements from the set that match "Y". Since you already removed all the "Y"s, there are no "Y"s to get back.
This isn't very "math like", but I believe this can be solved with this simple analogy.
Vaughan Hilts
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@BrennaRich Asaf has given a more mathematical explanation if you need it, just use the definitions to "unroll" it. – Vaughan Hilts Apr 07 '14 at 22:55
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$(X-Y)\cap Y$. What does it mean? It is the set of common elements between $Y$ and some set minus the elements of $Y$. The two sets are obviously disjoint, giving you the result: $(X-Y)\cap Y=\emptyset$.
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@BrennaRich See that this is an elementary consequence of the definition of intersection of sets and set subtraction. To see this geometrically, in the Euclidean plane, at least, draw a region. Call it $X$. From this, remove a region $Y$. Where do the regions $X$ and $Y$ intersect? – Apr 07 '14 at 22:57
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Sets being equal means that if an element $x$ belongs to the first, then it belongs to the second and vice versa.
To prove the given claim:
Take $x \in (X\setminus Y) \cap Y$. Then $x\notin Y$ and $x\in Y$. As we can see this is a contradiction, in other words you can imply falsity from these claims. From falsity ($f$), you can imply anything, especially that $x\in \emptyset$.
The other direction (from $x\in \emptyset$ to $x \in (X\setminus Y) \cap Y$) goes in a similar way.
This is rather informal way of stating the proof, but since this was about elementary set theory, I guess the proof would be something like this.
ploosu2
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