Show function equality is false

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Discussion Overview

The discussion revolves around demonstrating that the equality f(Y&Z) = f(Y) & f(Z) is false by providing examples of functions f: X -> A where this equality does not hold. The scope includes mathematical reasoning and exploration of function properties.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant seeks assistance in finding a counterexample to the equality f(Y&Z) = f(Y) & f(Z).
  • Another participant suggests trying simple finite sets and defining mappings in various ways to explore the problem.
  • A participant proposes a function f: (x,y) -> {x+1,y-1} and attempts to evaluate it, expressing confusion about the intersection of sets.
  • Another participant corrects the misunderstanding about the intersection of sets, clarifying that XnY is empty, not zero, and encourages constructing the function differently.
  • A suggestion is made to consider sets with overlapping elements and a specific mapping to explore the equality further.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the examples or the approach to take, with multiple competing views and ongoing confusion about the function definitions and set operations.

Contextual Notes

There are limitations in the understanding of set operations and function definitions among participants, particularly regarding the intersection of sets and the nature of the functions being constructed.

cateater2000
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Hi there I'm having trouble with this question, any tips would be great.

Show the following is false, by giving an example of a function(s) f:X->A for which the equalities fail:

f(Y&Z)=f(Y)&f(Z);


thanks for your help.
 
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try some examples and let us see what you come up with.

remember a function can be any way at all of mapping the elements of one set to the elements of another set. so start with some real simple finite sets and define the maps in various ways.
 
I came up with something like this but it doesn't seem right.

f: (x,y)->{x+1,y-1}

then i had something like
X={(1,2)}
y={(0,0)}

f(X) gives us (2,1)
f(Y) gives us (1,-1)
so f(X) & f(Y) is just (1)

But f(X&Y) I'm confused because X&Y = 0
I'm not sure if I'm on the right track or not, thanks again for the help.
 
XnY is empty not zero, they are different.
You're constructing the function so you can make it so that this problem goes away.

Suppose that X and Y are sets of two elements that have one element in common, say

{1,2} and {2,3}

what about it f were the map that takes 1 and 3 to 'a', and 2 to 'b' what happens there?
 

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