Which of the following quantified propositions is true if $ \ U=\mathcal{P} \{1,2,3 \} \ $ ?
(i) $ (\exists A)_{U} (\exists B)_U (|A|>|B| \ \ and \ \ A \cap B=\{1,2,3 \}) $
(ii) $ (\forall A)_{U} (\forall B)_U (|A|=|B| \ \ and \ \ A \cap B=\{1 \})$
(iii) $ (\exists A)_{U} (\forall B)_U (|A|=|B| \ \ and \ \ A \cap B=\{1 \})$
(iv) $ (\exists A)_{U} (\exists B)_U (|A|=|B| \ \ and \ \ A \cap B=\{1 \})$
(v) $ (\forall A)_{U} (\forall B)_U (|A|>|B| \ \ and \ \ A \cap B=\{1,2,3 \})$
Answer:
The power set $ \ U =\left\{\phi, \{1\}, \{2 \}, \{3 \}, \{1,2 \}, \{1,3 \}, \{2,3 \}, \{1,2,3 \}\right\} \ $
(i)True,
Take $ \ A=\{1,2,3 \} \ $ and $ \ B=\{1,2 \} \ $
Then $ \ |A|=3>2=|B| \ $ and $ \ A \cap B=\{1,2,3 \} \ $
(ii) False,
It can be seen taking $ \ A=\phi \ $
(iii) False,
It can be seen taking $ \ A=\phi \ $
(iv) True,
Take $ \ A=\{1 \}=B \ $
(v)
False,
Am I right?
If not correct me
i) $A\cap B = \{1,2,3\}$ Implies $\{1,2,3\} = U \subset B$. That can only happen if $U = B$. Likewise $U \subset A$ so $U = A = B$ and $|A| = |B|$.
So it is false that there exist any two subsets with these conditions.
ii) Obviously that can't be true for ALL $A$ and $B$. Just take two that are of different sizes; or that don't have $\{1\}$ as their intersections. The counterexamples are copious.
iii) This is saying there exists an $A$ so that for all $B$ the following are true. Well we take any $B$ that doesn't contain $1$ that won't be true for any $A$.
Note: $A =\emptyset$ is not a counterexample, as the statement is there exists an $A$ (which doesn't have to be $\emptyset$.) However $B = \emptyset$ would be a counter example as $A \cap \emptyset \ne \{1\}$ for any possible $A$.
iv) This is claiming there exist one $A$ and one $B$ with to property.
There are several examples. $A = B=\{1\}$ is the simplest but $A = \{1,2\}; B = \{1,3\}$ also works as does $B = \{1,2\}; A= \{1,3\}$.
v) For all possible $A$ and $B$. i) showed there weren't any, so it clearly isn't true for all. ANY $A$ and $B$ will be a counter-example.