Question

Let f be a function from A to B. Let S and T be subsets of B .

Show that \({f^{ - 1}}(S) = \overline {{f^{ - 1}}(S)} \)

Short answer

The function is

\({f^{ - 1}}(S) = \overline {{f^{ - 1}}(S)} \)

Step-by-step solution

Step 1:

Complement \(\overline A \): All elements in the universal set U NOT in A.

X is a subset of Y if every element of X is also an element of Y.

Notation:\(X \subseteq Y\)

Step 2:

Given: (a)

\(f:A \to B\)

\(S \subseteq B\)

To proof: \({f^{ - 1}}(S) = \overline {{f^{ - 1}}(S)} \)

PROOF

FIRST PART

Let \(x \in {f^{ - 1}}(\overline S )\) . then there exists an element \(y \in \overline S \) such that\(f(x) = y\).

By the definition of the complement:

\(y \in S\)

Since \(f(x) = y\)

\(f(x) \in S\)

\({f^{ - 1}}(S)\)contains all elements that are the image of all an element in S.

\(x \in \overline {{f^{ - 1}}(S)} \)

By the definition of union:

\(x \in {f^{ - 1}}(S) \cup {f^{ - 1}}(T)\)

By the definition of subset:

\({f^{ - 1}}(\overline S ) = \overline {{f^{ - 1}}(S)} \)

Step 3:

FIRST PART

Let \(x \in \overline {{f^{ - 1}}(S)} \) . then there exists an element \(y \in \overline S \) such that\(f(x) = y\).

By the definition of the complement:

\(x \in {f^{ - 1}}(S)\)

\({f^{ - 1}}(S)\)contains all elements that are the image of all an element in S.

\(x \in {f^{ - 1}}{(\overline S )^{}}\)

By the definition of subset

\[\overline {{f^{ - 1}}(S)} \subseteq {f^{ - 1}}(\overline S )\]

The two sets have to be equal:

\({f^{ - 1}}(\overline S ) = \overline {{f^{ - 1}}(S)} \)

Hence, the solution is

\({f^{ - 1}}(\overline S ) = \overline {{f^{ - 1}}(S)} \)