Snapsolve any problem by taking a picture.
Try it in the Numerade app?

This problem has been solved

You'll get a detailed solution from a subject matter expert that helps you learn core concepts.

Question

Suppose $f: A \rightarrow B$ and $C \subseteq A$. The set $f \cap(C \times B)$, which is a relation from $C$ to $B$, is called the restriction of $f$ to $C$, and is sometimes denoted $f \nmid C$. In other words, $$ f \mid C=f \cap(C \times B) $$ (a) Prove that $f \mid C$ is a function from $C$ to $B$ and that for all $c \in$ $C, f(c)=(f \backslash C)(c)$. (b) Suppose $g: C \rightarrow B$. Prove that $g=f \backslash C$ iff $g \subseteq f$. (c) Let $g$ and $h$ be the functions defined in parts 2 and 3 of Example 5.1.3. Show that $g=h \downarrow \mathbb{Z}$.

   Suppose $f: A \rightarrow B$ and $C \subseteq A$. The set $f \cap(C \times B)$, which is a relation from $C$ to $B$, is called the restriction of $f$ to $C$, and is sometimes denoted $f \nmid C$. In other words,
$$
f \mid C=f \cap(C \times B)
$$
(a) Prove that $f \mid C$ is a function from $C$ to $B$ and that for all $c \in$ $C, f(c)=(f \backslash C)(c)$.
(b) Suppose $g: C \rightarrow B$. Prove that $g=f \backslash C$ iff $g \subseteq f$.
(c) Let $g$ and $h$ be the functions defined in parts 2 and 3 of Example 5.1.3. Show that $g=h \downarrow \mathbb{Z}$.

Show more…

How to Prove It: A Structured Approach

How to Prove It: A Structured Approach

Daniel J. Velleman 1st Edition

Chapter 5, Problem 5

Instant Answer

Step 1

First, we know that $f$ is a function from $A$ to $B$, so for each $a \in A$, there exists a unique $b \in B$ such that $(a, b) \in f$. Since $C \subseteq A$, for each $c \in C$, there exists a unique $b \in B$ such that $(c, b) \in f$. Therefore, $f \mid C$ is Show more…

Show all steps

lock

Mengchun Cai and 58 other educators are ready to help you.

Ask a new question

Recommended Videos

Transcript

00:02 Here we are given a function f which goes from a to b and we are given two subsets of b, let's call them t and s.

00:13 We are first required to prove the preimage of the union will be equal to the union of the preimage.

00:23 Okay, now let's first consider for any x which is contained in the left set, that means we can find some, there exists some y in this set, in this union such that fx is equal to y.

00:51 Okay, now we have two cases, we need to consider two cases.

00:54 First, suppose y now is in s, that means, suppose y is in s, that means fx is in s.

01:09 This tells us x must be contained in the preimage of s.

01:19 Okay, then this is actually a subset of the union.

01:26 So in the first case, x is contained in the right side.

01:33 Second, now assume y is in t, then we know fx is contained in t.

01:40 This tells us now x is contained in the preimage of t.

01:47 Again, this is a subset of the union.

01:53 So combining those two cases, we know for any x in this set, fx must be contained in s or t.

02:04 That means this preimage of the union is a subset of the union for the preimage.

02:17 Okay, now let's prove, now we want to prove this direction.

02:24 Okay, for any x now which is contained in the union, now we also need to consider two cases first.

02:37 Let's assume x is contained in the first set.

02:42 That means there exists some y which is contained in s, such that fx is in y which is contained in s.

02:57 Okay, now we notice s as a set is a subset of s union t.

03:10 We have this relationship.

03:12 That means if x is contained in the preimage of s, then there must exist some y which is contained in s, and which is also contained in a larger set, such that fx is equal to some element in s union t.

03:31 That means x must be contained.

03:45 Okay, now let's suppose x is in f inverse t.

03:51 And from the same logic, we know there is some y in t.

03:55 And as y is in t, y is also contained in this set, such that fx is equal to y which is contained in, and it is also contained in this union...

Previous Question