Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $A$, $B$, and $C$ are sets and $B \subseteq C$, how would I be able to show that $(A - B) - C = A - C$? I understand how this would be trivial if the set $B$ has no elements (or in other words its basically empty), but I'm confused as to how would you go about proving this otherwise?

share|improve this question

3 Answers 3

up vote 2 down vote accepted
  1. Suppose $x\in(A-B)-C$. Then we know $x\in A-B$ and $x\notin C$. From $x\in A-B$ we know $x\in A$ and $x\notin B$. Since we know $x\in A$ and $x\notin C$, we know $x\in A-C$.

  2. Suppose $x\in A-C$. Then we know $x\in A$ and $x\notin C$. Since $x\notin C$ and $B\subseteq C$, we know $x\notin B$. Then since $x\in A$ and $x\notin B$, we know $x\in A-B$. Since $x\in A-B$ and $x \notin C$, we know $x\in (A-B)-C$.

From (1), we know $(A-B)-C \subseteq A-C$.

From (2), we know $A-C \subseteq (A-B)-C$.

Therefore, $(A-B)-C = A-C$.

share|improve this answer

$$\begin{align}&x\in(A\setminus B)\setminus C\\ \iff &x\in A\setminus B\land x\notin C\\ \iff &x\in A\land x\notin B\land x\notin C\end{align}$$ From $B\subseteq C$ we have $x\in B\rightarrow x\in C$ or equivalently via propositional logic $x\notin C\rightarrow x\notin B$ and $x\notin C\leftrightarrow (x\notin B\land x \notin C)$. Hence ultimately $$\begin{align}&x\in(A\setminus B)\setminus C\\ \iff &x\in A\land x\notin C\\ \iff &x\in A\setminus C \end{align}$$

share|improve this answer

In elementary problems of this type you should think first of proving the equality of the two sets by showing that each is a subset of the other: $(A\setminus B)\setminus C\subseteq A\setminus C$ and $A\setminus C\subseteq(A\setminus B)\setminus C$. Very often the easiest way to do this is by ‘element-chasing’: to show that $X\subseteq Y$, let $x$ be an arbitrary element of $X$ and show somehow that this forces $x$ to belong to $Y$.

In your problem, to show that $(A\setminus B)\setminus C\subseteq A\setminus C$ you might start like this:

Let $x\in(A\setminus B)\setminus C$ be arbitrary; we wish to show that $x\in A\setminus C$, i.e., that $x\in A$ and $x\notin C$. By hypothesis $x\in A\setminus B$ and $x\notin C$, so all that remains is to show that $x\in A$. But that’s clear: $x\in A\setminus B$, so $x\in A$ and $x\notin B$. We don’t care that $x\notin B$: we now know that $x\in A$ and $x\notin C$, and hence that $x\in A\setminus C$, as desired.

I’ll leave the other inclusion to you; try to follow the same general scheme, and don’t be afraid to use words as well as symbols, just as I did above. The idea is to produce an argument that is both convincing and clear to the reader.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.