Let \(\left(f_{i}ight)_{i \in I}\) be arbitrarily many maps \(f_{i}: X ightarrow \mathbb{R}\). Show that

(i) \(\left\{\sup _{i} f_{i}>\lambdaight\}=\bigcup_{i}\left\{f_{i}>\lambdaight\}\);

(iii) \(\left\{\sup _{i} f_{i} \geqslant \lambdaight\} \supset \bigcup_{i}\left\{f_{i} \geqslant \lambdaight\}\)

(ii) \(\left\{\sup _{i} f_{i}<\lambdaight\} \subset \bigcap_{i}\left\{f_{i}<\lambdaight\}\);

(v) \(\left\{\inf _{i} f_{i}>\lambdaight\} \subset \bigcap_{i}\left\{f_{i}>\lambdaight\}\);

(iv) \(\left\{\sup _{i} f_{i} \leqslant \lambdaight\}=\bigcap_{i}\left\{f_{i} \leqslant \lambdaight\}\)

(vii) \(\left\{\inf _{i} f_{i} \geqslant \lambdaight\}=\bigcap_{i}\left\{f_{i} \geqslant \lambdaight\}\);

(vi) \(\quad\left\{\inf _{i} f_{i}<\lambdaight\}=\bigcup_{i}\left\{f_{i}<\lambdaight\}\)

(viii) \(\quad\left\{\inf _{i} f_{i} \leqslant \lambdaight\} \supset \bigcup_{i}\left\{f_{i} \leqslant \lambdaight\}\).