For each n in N, let A_n = {(n+1)k mid k in N | vedclass

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(B) Given that $A_n = \{(n+1)k \mid k \in N\}$.For $n=1$, $A_1 = \{2k \mid k \in N\} = \{2, 4, 6, 8, \dots\}$.For $n=2$, $A_2 = \{3k \mid k \in N\} =

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one-one but not onto

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(B) Given that $A_n = \{(n+1)k \mid k \in N\}$.For $n=1$, $A_1 = \{2k \mid k \in N\} = \{2, 4, 6, 8, \dots\}$.For $n=2$, $A_2 = \{3k \mid k \in N\} = \{3, 6, 9, 12, \dots\}$.For $n=3$, $A_3 = \{4k \mid k \in N\} = \{4, 8, 12, 16, \dots\}$.Now, $X = \bigcup_{n \in N} A_n = A_1 \cup A_2 \cup A_3 \cup \dots = \{2, 3, 4, 5, 6, \dots\}$.Here, $f: X ightarrow N$ is defined by $f(x) = x$ for all $x \in X$.Since $f(x) = x$ is an identity function on the domain $X$, it is clearly one-one.However, the codomain is $N = \{1, 2, 3, 4, \dots\}$.For $f$ to be onto, the range must equal the codomain. Here, the range is $X = \{2, 3, 4, 5, \dots\}$.Since $1 \in N$ but $1 
otin X$, there is no $x \in X$ such that $f(x) = 1$.Therefore, $f$ is not onto.

For each $n \in N$, let $A_n = \{(n+1)k \mid k \in N\}$ and $X = \bigcup_{n \in N} A_n$. $A$ mapping $f: X \rightarrow N$ defined by $f(x) = x, \forall x \in X$, is

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