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

[cbarker1]

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

 

[Euge]

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 $A\subset B\cap C$.