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Chapter 1: Problem 12

Show that if \(f: A \rightarrow B\) and \(E, F\) are subsets of \(A\), then \(f(E \cup F)=f(E) \cup f(F)\) and \(f(E \cap F) \subseteq f(E) \cap f(F)\)

Short Answer

Expert verified
The aforementioned steps show that \(f(E \cup F) = f(E) \cup f(F)\) and \(f(E \cap F) \subseteq f(E) \cap f(F)\). These are properties of set mappings under a function f.

Step by step solution

01

Starting with the first proof

We will begin by showing \(f(E \cup F) = f(E) \cup f(F)\). To show this, we must prove two things: that \(f(E 鈭� F)\) is a subset of \(f(E) 鈭� f(F)\) and vice versa.
02

Prove \(f(E \cup F) \subseteq f(E) \cup f(F)\)

Pick an arbitrary \(y\) from \(f(E \cup F)\). By definition of the image under a function, there exists an element \(x\) in \(E \cup F\) such that \(f(x) = y\). Now, since \(x\) is in \(E \cup F\), \(x\) is in \(E\) or \(x\) is in \(F\). Thus, \(y = f(x)\) is in \(f(E)\) or in \(f(F)\). In other words, \(y\) is in \(f(E) \cup f(F)\). Therefore, \(f(E 鈭� F) 鈯� f(E) 鈭� f(F)\).
03

Prove \(f(E \cup F) \supseteq f(E) \cup f(F)\)

Now, pick an arbitrary \(y\) from \(f(E) \cup f(F)\). This implies that \(y\) is in \(f(E)\) or \(y\) is in \(f(F)\). Then there exists \(x\) in \(E\) or \(F\) such that \(f(x) = y\). But \(x\) is then in \(E \cup F\), and so \(y = f(x)\) is in \(f(E \cup F)\). Therefore, \(f(E \cup F) \supseteq f(E) \cup f(F)\). Hence, \(f(E 鈭� F) = f(E) 鈭� f(F)\).
04

Next up - intersection proof

Next, we need to show that \(f(E \cap F) \subseteq f(E) \cap f(F)\). We do not have equality in general - only the subset relation.
05

Prove \(f(E \cap F) \subseteq f(E) \cap f(F)\)

Take an arbitrary \(y\) from \(f(E \cap F)\). By definition, there exists an element \(x\) in \(E \cap F\) such that \(f(x) = y\). Now, \$x in \(E \cap F\) means \(x\) is in both \(E\) and \(F\). Thus, \(y = f(x)\) is in both \(f(E)\) and \(f(F)\). In other words, \(y\) is in \(f(E) \cap f(F)\). Therefore, \(f(E \cap F) \subseteq f(E) \cap f(F)\).

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