# Elementary set theory relation

If $U = A \backslash B$ , do we then have $A \backslash U = B$ or just $A \backslash U \subseteq B$?

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HINT: Consider the case that $B$ is not a subset of $A$.

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ah, so the answer is no, thank you. – user66906 May 5 '14 at 22:28

Hint: To understand what is $A\setminus U$ you need to understand what elements of $A$ are also in $U$.

Those are exactly the elements that are not also in $B$, what are we left with ?

I suggest you draw a venn diagram to see things more clearly

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We have that $$X\setminus Y = X \cap Y^c.$$

Using this identity your question is:

$$\text{Does }U = A \cap B^c\text{ imply } A\cap U^c = B \text{ ?}$$

Let's see:

\begin{align} A \cap U^c &= A \cap (A \cap B^c)^c \\ &= A \cap(A^c \cup B) \\ &= (A \cap A^c) \cup (A \cap B) \\ &= \varnothing \cup (A \cap B) = A \cap B \end{align}

Surely $A \cap B \subseteq B$, but without any assumptions $A \cap B \neq B$. The last part is true if and only if $B \subseteq A$, so to find a counterexample, take $A = \varnothing$ and $B = \{\spadesuit\}$. Now, the premise tells us that $U = A \setminus B = \varnothing$, but $A \setminus U = \varnothing \neq \{\spadesuit\} = B$.

I hope this helps $\ddot\smile$

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