Topology Munkres Chapter 1 exercise 2 e- Set theory

[cbarker1]

Dear Everyone
I am having some difficulties on exercise 2e from Topology 2nd ed by J. Munkres . Here are the directions:
determine which of the following states are true for all sets [FONT=MathJax_Math-italic]A, [FONT=MathJax_Math-italic]B, [FONT=MathJax_Math-italic]C, and [FONT=MathJax_Math-italic]D. If a double implication fails, determine whether one or the other one of the possible implication holds. If an equality fails, determine whether the statement becomes true if the "equal" symbol is replaced by one or the other of the inclusion symbols [FONT=MathJax_Main]⊂or [FONT=MathJax_Main]⊃.e.) $A-(A-B)=B$

My work:
Let A={1,2,3,4,5,7} and B={1,3,5,9}
A-B={2,4,7}
A-(A-B)=A-{2,4,7}={1,3,5,9}= B?

Thanks
Cbarker1

 

[Opalg]

Dear Everyone
I am having some difficulties on exercise 2e from Topology 2nd ed by J. Munkres . Here are the directions:
determine which of the following states are true for all sets [FONT=MathJax_Math-italic]A, [FONT=MathJax_Math-italic]B, [FONT=MathJax_Math-italic]C, and [FONT=MathJax_Math-italic]D. If a double implication fails, determine whether one or the other one of the possible implication holds. If an equality fails, determine whether the statement becomes true if the "equal" symbol is replaced by one or the other of the inclusion symbols [FONT=MathJax_Main]⊂or [FONT=MathJax_Main]⊃.e.) $A-(A-B)=B$

My work:
Let A={1,2,3,4,5,7} and B={1,3,5,9}
A-B={2,4,7}
A-(A-B)=A-{2,4,7}={1,3,5,9}= B?

Thanks
Cbarker1

If A={1,2,3,4,5,7} and B={1,3,5,9} then
A-B={2,4,7}
A-(A-B)=A-{2,4,7}={1,2,3,4,5,7}-{2,4,7}={1,3,5}.

Notice that A-(A-B) is contained in B, but it is not the whole of B because it does not include 9.

That shows that the statement $A-(A-B)=B$ is not true. However, it is true that $A-(A-B)\subset B$. Can you prove that?

 

[cbarker1]

If A={1,2,3,4,5,7} and B={1,3,5,9} then
A-B={2,4,7}
A-(A-B)=A-{2,4,7}={1,2,3,4,5,7}-{2,4,7}={1,3,5}.

Notice that A-(A-B) is contained in B, but it is not the whole of B because it does not include 9.

That shows that the statement $A-(A-B)=B$ is not true. However, it is true that $A-(A-B)\subset B$. Can you prove that?

Proof: Suppose x is an arbitrary element in $A-(A-B)$. Then $x\in A$ and $x \notin(A-B)$. So $x\in A$ and $x\notin A$ and $x\in B$. Thus $x\in B$ because an element $x$ can't be both in A and not in A. QED

 

[Evgeny.Makarov]

Proof: Suppose x is an arbitrary element in $A-(A-B)$. Then $x\in A$ and $x \notin(A-B)$. So $x\in A$ and $x\notin A$ and $x\in B$.

The fact that $x\in A-B$ means that $x\in A$ and $x\notin B$. The negation of that, i.e., $x\notin A-B$, means that $x\notin A$ or $x\in B$. Thus, writing $\land$ for "and" and $\lor$ for "or", $x\in A-(A-B)$ means
$$
\begin{align}
x\in A\land (x\notin A\lor x\in B)&\iff (x\in A\land x\notin A)\lor (x\in A\land x\in B)\\
&\iff x\in A\land x\in B\\
&\implies x\in B
\end{align}
$$
In particular, we deduced that $A-(A-B)=A\cap B$.