1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof involving functions

  1. Oct 27, 2007 #1
    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.
     
  2. jcsd
  3. Oct 27, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    What, exactly, is your definition of [itex] f \cap g [/itex]?
     
  4. Oct 27, 2007 #3
    I'm guessing just the normal definition of intersection of sets.

    All the ordered pairs that are common to both f and g.
     
  5. Oct 27, 2007 #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).
     
  6. Oct 27, 2007 #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.
     
  7. Oct 27, 2007 #6

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Unless you're working in an algebra of relations, in which case the operation is fairly natural.
     
  8. Oct 27, 2007 #7
    Well there you go: it's not true.

    Unions are functions when the domains are disjoint.
     
  9. Oct 27, 2007 #8
    i'm always afraid to write a "this is wrong"

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

    Scares me.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Proof involving functions
  1. Proofs involving subsets (Replies: 22)

  2. Function proof (Replies: 3)

Loading...