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Today , my professor has written a set formula which I don't fully understand: $$ A = (A\setminus B) \cup B $$ Please help me to fill in the missing details for the above, and its proof. My notes are incomplete.

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    $\begingroup$ The claim is true if and only if $A \cap B = B$ (B subset of or equal to A). $\endgroup$ – Namaste Sep 11 '13 at 18:20
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    $\begingroup$ @amWhy. Why not turn this into an answer (with a bit of explanation)? $\endgroup$ – Rick Decker Sep 11 '13 at 19:25
  • $\begingroup$ @RickDecker Done! $\endgroup$ – Namaste Sep 11 '13 at 19:31
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Others have given examples why your posted statement is not always true.

I'll expand on a comment I left earlier. IF one of the premises we are working with is that $B \subseteq A$, then the claim does in fact hold.

In fact claim is true if and only if $A\cap B = B$ (which is equivalent to the premise $B\subseteq A$). The important thing here is whether you see why $$B \subseteq A \implies A = (A \setminus B) \cup B\quad?$$

If we have that $B\subseteq A$, and perform $A \setminus B$, then by definition of set-minus, we take out of $A$ all the elements in $B$ (where $B\subseteq A$), then when we perform $(A\setminus B) \cup B$, by definition of set union, we are putting back all the previously removed elements back into A, leaving us with $A$ itself.

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  • $\begingroup$ Thanks a lot, that's explanation totally makes sense for the proof my prof was working on. I missed that minor detail about B⊆A, thus he was talking a little bit fast. $\endgroup$ – Peter Sep 12 '13 at 1:04
  • $\begingroup$ "That is" is misleading since, before it there is an implication while after it there is an equivalence. Both are true but the latter cannot be an explanation of the former. $\endgroup$ – Did Sep 16 '13 at 5:55
  • $\begingroup$ You're correct, @Did. I deleted the misleading words. Thank you. $\endgroup$ – Namaste Sep 16 '13 at 12:03
  • $\begingroup$ @Amzoti What is a TU? $\endgroup$ – Ahaan S. Rungta Nov 28 '13 at 18:40
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    $\begingroup$ @Ahaan: A "TU" simply means a "Thumb's up", as in an "up vote"! ;-) $\endgroup$ – Namaste Nov 28 '13 at 18:54
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The claim's false. For example, take

$$A=\{1,2\}\;,\;\;B=\{1,3\}\implies A=\{1,2\}\neq \{2\}\cup\{1,3\}=\{1,2,3\}=(A\setminus B)\cup B\;.$$

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$A\setminus B = \left\{x \mid x \in A\text{ and }x\not\in B\right\}$

$A\cup B = \left\{x \mid x \in A\text{ or }\in B\right\}$


By the way this formula false.

Take $A=\emptyset$ and $B\not= \emptyset$ (for example $\left\{0\right\}$. Then $\left(A\setminus B\right) = \emptyset$ and so $\left(A\setminus B\right)\cup B = B\not= \emptyset = A$.

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$A\setminus B$ is the set of things that are in $A$ but not in $B$. $X\cup Y$ is the set of things that are in $X$, in $Y$, or in both $X$ and $Y$. Thus, the set $(A\setminus B)\cup B$ consists of those things that are in $A$ but not in $B$, and those things that are in $B$. (There are no things in both $A\setminus B$ and $B$: anything in $A\setminus B$ is by definition not in $B$.) The equation

$$A=(A\setminus B)\cup B$$

therefore says that the members of $A$ are the things that are in $A$ but not in $B$, and the things that are in $B$. In particular, this says that everything in $B$ belongs to $A$. This certainly isn’t always true, and the other answers have given you some specific counterexamples.

What is true is that

$$A=(A\setminus B)\cup(A\cap B)\;.$$

$A\cap B$ is the set of things belonging to both $A$ and $B$, so this just says that $A$ consists of those things that are in $A$ but not $B$ and those things that are in $A$ and in $B$. If $a\in A$, then either $a\in B$, in which case $a\in A\cap B$, or $a\notin B$, in which case $a\in A\setminus B$. Conversely, if $a\in(A\setminus B)\cup(A\cap B)$, then either $a\in A\setminus B$, or $a\in A\cap B$, and in both cases $a\in A$.

Note, by the way, that if $B\subset A$, then $A\cap B=B$, and in this special case we do have

$$A=(A\setminus B)\cup(A\cap B)=(A\setminus B)\cup B\;.$$

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I think it's the topic is $A=(A\cap B)\cup B$ or $A=(A\cup B)\cap B$ (note: this is inequatily modulus)

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