MHB Topology Munkres Chapter 1 exercise 2 b and c- Set theory equivalent statements

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Exercise 2b from Munkres' topology discusses the relationship between set inclusions and unions, concluding that the implication holds true in both directions. A counterexample using rational numbers shows that while \( A \subset (B \cup C) \) can be true, \( A \subset B \) or \( A \subset C \) may not hold. Exercise 2c examines the relationship between set inclusions and intersections, confirming that if \( A \subset B \) and \( A \subset C \), then \( A \subset (B \cap C) \) is also true. The discussion highlights the importance of understanding implications and counterexamples in set theory. Overall, both exercises illustrate key concepts in set inclusion and their logical relationships.
cbarker1
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Dear Every one,

I am having some difficulties on exercise 2b and 2c from Topology 2nd ed by J. Munkres . Here are the directions:
determine which of the following states are true for all sets $A$, $B$, $C$, and $D$. 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 $\subset$ or $\supset$.

Problem 2b and 2c, respectively:

b. $A\subset B$ or $A\subset C \iff A \subset (B \cup C)$
c. $A\subset B$ and $A\subset C \iff A \subset (B\cap C)$

My attempt
Let $A=\left\{1,2,3,4,5\right\}$, $B=\left\{3,4,5\right\}$ and $C=\left\{1,2\right\}$.
b. I believe it to be true in $\implies$ and true in $\Longleftarrow$.
c. I believe it to be true in $\Longleftarrow$ and false in $\implies$.

Any counterexample can help me figure out why these are correct or incorrect.

thanks,
Cbarker1
 
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Hi Cbarker1,

In b., take $A$ to be the rationals, $B$ to be the interval $(-\infty, 0]$ and $C$ to be the interval $(0, \infty)$. Then $B \cup C$ is the reals so that $A \subset B \cup C$, but neither $A\subset B$ nor $A\subset C$. In c., if $a\in A$, then since $A \subset B$, $a\in B$; similarly, as $A\subset C$, $a\in C$. Hence $a\in B\cap C$. Since $a$ was arbitrary, one concludes $A\subset B\cap C$.
 
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