Proof involving functions


by Jacobpm64
Tags: functions, involving, proof
Jacobpm64
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#1
Oct27-07, 10:31 AM
P: 240
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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HallsofIvy
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#2
Oct27-07, 03:00 PM
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What, exactly, is your definition of [itex] f \cap g [/itex]?
Jacobpm64
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#3
Oct27-07, 03:03 PM
P: 240
I'm guessing just the normal definition of intersection of sets.

All the ordered pairs that are common to both f and g.

ZioX
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#4
Oct27-07, 04:12 PM
P: 372

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).
Jacobpm64
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#5
Oct27-07, 04:20 PM
P: 240
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.
Hurkyl
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#6
Oct27-07, 04:27 PM
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Quote Quote by ZioX View Post
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.
ZioX
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#7
Oct27-07, 04:33 PM
P: 372
Well there you go: it's not true.

Unions are functions when the domains are disjoint.
Jacobpm64
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#8
Oct27-07, 04:35 PM
P: 240
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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