Proving the Intersection of Functions

In summary, we are trying to prove that if f: A \rightarrow B and g: C \rightarrow D, then the intersection of f and g, f \cap g: A \cap C \rightarrow B \cap D, is a function. We begin by assuming that A, B, C, and D are sets and f and g are functions from A to B and C to D, respectively. We then take an arbitrary element a from A and show that there is an element b in B such that (a, b) is in f. Similarly, for an arbitrary element c in C, there is an element d in D such that (c, d) is in g. We must consider two cases: when
  • #1
Jacobpm64
239
0
Prove the following:

If [tex] f : A \rightarrow B [/tex] and [tex] g : C \rightarrow D [/tex], then [tex] f \cap g : A \cap C \rightarrow B \cap D[/tex].

Here's my thoughts/attempt:

Proof:
Let A, B, C, and D be sets. Assume [tex] f : A \rightarrow B [/tex] and [tex] g : C \rightarrow D [/tex]. Let [tex] a \in A [/tex]. Since f is a function from A to B, there is some [tex] y \in B [/tex] such that [tex] (a, y) \in f [/tex]. Let [tex] b \in B [/tex] be such an element, that is, let [tex] b \in B [/tex] such that [tex] (a,b) \in f [/tex]. Let [tex] c \in C [/tex]. Since g is a function from C to D, there is some [tex] z \in D [/tex] such that [tex] (c, z) \in g [/tex]. Let [tex] d \in D [/tex] be such an element, that is, let [tex] d \in D [/tex] such that [tex] (c,d) \in g [/tex].



This is all I have so far.

Would I have to break it into cases where [tex] a = c [/tex] and [tex] a \not= c [/tex]? If [tex] a = c [/tex], [tex] A \cap C [/tex] contains an element, but if [tex] a \not= c [/tex], [tex] A \cap C [/tex] is empty since a and c were arbitrary. The same argument holds for [tex] B \cap D [/tex]. So, taking these things into account, [tex] f \cap g [/tex] is either a function from the set containing a to the set containing b, or its a function from the empty set to the empty set.

Does this make any sense, is it necessary, and how should I write it in my proof?

Thanks in advance.
 
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  • #2
What, exactly, is your definition of [itex] f \cap g [/itex]?
 
  • #3
I'm guessing just the normal definition of intersection of sets.

All the ordered pairs that are common to both f and g.
 
  • #4
Just talk in terms of set theoretics. f is a set, g is a set. Show that the intersection of f and g defines a new relation on (A intersect C)x(B intersect D) that satisfies the definition of a function. (for every x in A intersect C there is some y in B intersect D such that (x,y) in the relation and that for any x in A intersect C this y is unique.)

What happens if we take unions? Do we still get a new function?

Also, no one really talks about functions this way (intersections and unions).
 
  • #5
Well, one of our earlier assignments was to disprove the case when we took unions.

So I know that you don't get a function when you take f union g.



I'm still not convinced that the intersection claim is correct though.

Earlier, on another forum, someone came up with the counterexample:
[tex]A = \left\{ {1,2,4} \right\}\,,\,B = \left\{ {p.q,r} \right\}\,,\,C = \left\{ {2,4,6} \right\}\,\& \,D = \left\{ {r,s,t} \right\}[/tex]
[tex]f:A \mapsto B,\quad f = \left\{ {(1,p),(2,r),(4,q)} \right\}[/tex]
[tex]g:C \mapsto D,\quad g = \left\{ {(2,r),(4,t),(6,s)} \right\}[/tex]

But [tex]f \cap g = \left\{ {(2,r)} \right\}[/tex] while [tex]A \cap C = \left\{ {2,4} \right\}[/tex] clearly [tex]f \cap g:A \cap C \not{\mapsto} B \cap D[/tex]
There is no mapping for the term 4.
 
  • #6
ZioX said:
Also, no one really talks about functions this way (intersections and unions).
Unless you're working in an algebra of relations, in which case the operation is fairly natural.
 
  • #7
Well there you go: it's not true.

Unions are functions when the domains are disjoint.
 
  • #8
i'm always afraid to write a "this is wrong"

on a homework assignment that says "prove the following theorem"

Scares me.
 

1. What is a function in mathematics?

A function in mathematics is a relationship between two sets of elements, where each element in the first set is paired with exactly one element in the second set. The first set is called the domain and the second set is called the range. Functions are commonly represented using notation such as f(x) or y = f(x).

2. What is a proof involving functions?

A proof involving functions is a method used to demonstrate that a mathematical statement involving functions is true. This usually requires logical reasoning and the use of mathematical properties and definitions to show that the statement is valid.

3. What are some common techniques used in proofs involving functions?

Some common techniques used in proofs involving functions include direct proof, proof by contradiction, proof by induction, and proof by cases. These techniques involve breaking down the problem into smaller, more manageable steps and using logic to show that each step is true.

4. How do I know when a function is invertible?

A function is invertible if it satisfies the one-to-one and onto properties. This means that each element in the domain is paired with exactly one element in the range, and every element in the range has at least one preimage in the domain. To prove that a function is invertible, you can use the horizontal line test or show that the function has an inverse function defined algebraically.

5. Can I use graphs to prove statements about functions?

Yes, graphs can be a useful tool in proving statements about functions. Graphical representations of functions can help to illustrate the relationship between the domain and range, and can also be used to visualize the effects of applying different operations to a function. However, it is important to supplement graphical evidence with logical reasoning and mathematical properties in order to provide a complete proof.

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