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Topology Munkres Chapter 1 exercise 2 e- Set theory

Cbarker1

Active member
Jan 8, 2013
236
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], [FONT=MathJax_Math-italic]B[/FONT], [FONT=MathJax_Math-italic]C[/FONT], and [FONT=MathJax_Math-italic]D[/FONT]. 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]⊂[/FONT]or [FONT=MathJax_Main]⊃[/FONT].


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

MHB Oldtimer
Staff member
Feb 7, 2012
2,679
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], [FONT=MathJax_Math-italic]B[/FONT], [FONT=MathJax_Math-italic]C[/FONT], and [FONT=MathJax_Math-italic]D[/FONT]. 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]⊂[/FONT]or [FONT=MathJax_Main]⊃[/FONT].


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

Active member
Jan 8, 2013
236
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

Well-known member
MHB Math Scholar
Jan 30, 2012
2,488
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
\(\displaystyle
\begin{align}
x\in A\land (x\notin A\lor x\in B)&\iff (x\in A\land x\notin A)\lor (x\in A\land x\in B)\\
&\iff x\in A\land x\in B\\
&\implies x\in B
\end{align}
\)
In particular, we deduced that $A-(A-B)=A\cap B$.