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Proof involving functions 
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#1
Oct2707, 10:31 AM

P: 239

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
Oct2707, 03:00 PM

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What, exactly, is your definition of [itex] f \cap g [/itex]?



#3
Oct2707, 03:03 PM

P: 239

I'm guessing just the normal definition of intersection of sets.
All the ordered pairs that are common to both f and g. 


#4
Oct2707, 04:12 PM

P: 371

Proof involving functions
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
Oct2707, 04:20 PM

P: 239

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
Oct2707, 04:27 PM

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#7
Oct2707, 04:33 PM

P: 371

Well there you go: it's not true.
Unions are functions when the domains are disjoint. 


#8
Oct2707, 04:35 PM

P: 239

i'm always afraid to write a "this is wrong"
on a homework assignment that says "prove the following theorem" Scares me. 


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