91Ó°ÊÓ

We can define an injective function \(f: A \to \mathcal{P}(A)\) by assigning each element \(a \in A\) to a unique subset of \(A\), which is the singleton set containing \(a\): $$ f(a) = \{a\} $$ Now, we need to show that this function is injective. Let \(a, b \in A\) with \(a \neq b\). By the definition of the function, $$ f(a) = \{a\} \text{ and } f(b) = \{b\} $$ Since \(a \neq b\), we have that \(\{a\} \neq \{b\}\), which means \(f(a) \neq f(b)\). Therefore, the function is injective.

To show that there is no surjective mapping from \(A\) to \(\mathcal{P}(A)\), we will use proof by contradiction. Assume that there exists a function \(f: A \to \mathcal{P}(A)\) that is surjective. This means that, for every subset \(B \subseteq A\), there exists an element \(a \in A\) such that \(f(a) = B\). Now, define a subset \(C \subseteq A\) as follows: $$ C = \{a \in A \mid a \notin f(a)\} $$ Since \(f\) is assumed to be surjective, there must exist an element \(x \in A\) such that \(f(x) = C\). Now we consider two cases: 1. \(x \in C\): According to the definition of set \(C\), if \(x \in C\), then \(x \notin f(x)\). But we know that \(f(x) = C\), so \(x \notin C\). This leads to a contradiction. 2. \(x \notin C\): In this case, according to the definition of the set \(C\), we have \(x \notin C\), or equivalently, \(x \in f(x)\). But we know that \(f(x) = C\), so \(x \in C\). This also leads to a contradiction. In both cases, we arrived at contradictions. Thus, our assumption that there exists a surjective function \(f: A \to \mathcal{P}(A)\) is false, and we have proven that there is no surjective mapping from \(A\) to \(\mathcal{P}(A)\).