Let A = {0, {Φ}, 1, {1, Φ}, 2}
Let \[ A=\{0,\{\Phi\},1,\{1,\Phi\},2\} \] Which of the following are true?
(i) \(\Phi \in A\)
(ii) \(\{\Phi\} \in A\)
(iii) \(\{1\} \in A\)
(iv) \(\{2,0\} \subset A\)
(v) \(2 \subset A\)
(vi) \(\{2,\{1\}\} \nsubseteq A\)
(vii) \(\{\{2\},\{1\}\} \nsubseteq A\)
(viii) \(\{0,\{\Phi\},\{1,\Phi\}\} \subset A\)
(ix) \(\{\{\Phi\}\} \subset A\)
Solution
Elements of \[ A \] are \[ 0,\ \{\Phi\},\ 1,\ \{1,\Phi\},\ 2 \]
(i) False, because \[ \Phi \notin A \]
(ii) True, because \[ \{\Phi\} \in A \]
(iii) False, because \[ \{1\} \notin A \]
(iv) True, because \[ 2,0 \in A \]
(v) False, because \[ 2 \] is not a set.
(vi) True, because \[ \{1\} \notin A \] Hence \[ \{2,\{1\}\} \nsubseteq A \]
(vii) True, because \[ \{2\},\{1\} \notin A \]
(viii) True, because all elements belong to \[ A \]
(ix) True, because \[ \{\Phi\} \in A \]