Functions, intersections, and unions

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Homework Help Overview

The discussion revolves around demonstrating set relationships involving a function g and the operations of intersection and union on sets A and B. The original poster seeks to establish that g(A ∩ B) is a subset of g(A) ∩ g(B) and is also looking for guidance on a similar relationship for unions.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the proof of the subset relationship for intersection and explore the implications of set properties. There is an inquiry into the relationship for unions, with suggestions on how to approach it. The original poster also seeks clarification on the names of certain set properties.

Discussion Status

Some participants have provided insights into the proof for the intersection case and have suggested a possible direction for the union case. The original poster has expressed uncertainty about the union relationship and is encouraged to explore it further. There is an ongoing exploration of set properties and their names.

Contextual Notes

The original poster is working within the constraints of a homework assignment that requires them to derive relationships without being given complete solutions. There is a mention of properties learned in a previous class, indicating some prior knowledge but also a gap in recall.

lokisapocalypse
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Okay I need to show that for a function g,

g(A n B) is a subset of g(A) n g(B) where A and B are sets and n means intersection.

I also need to show something similar about the union but I'm not given the relation. I have to figure that out myself. Any pointers on how to get started on either of these?
 
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To show that:

[tex]g(A \cap B) \subseteq g(A) \cap g(B)[/tex]

it suffices to show that:

[tex]x \in g(A \cap B) \Rightarrow x \in g(A) \cap g(B)[/tex]

This basically says that if you have any sets X and Y, and X is inside Y (i.e. X is a subset of Y), then any element inside X must obviously be inside Y. So, start by assuming [itex]x \in g(A \cap B)[/itex] and deduce [itex]g(A) \cap g(B)[/itex]. It's pretty clear. If [itex]x \in g(A \cap B)[/itex], then [itex]x \in g(A)[/itex], since [itex](A \cap B) \subseteq A[/itex]. Similarly, if [itex]x \in g(A \cap B)[/itex], then [itex]x \in g(A)[/itex]. So we have that if x is in [itex]g(A \cap B)[/itex], then x is in both g(A) and g(B). Obviously, this means that x is in [itex]g(A) \cap g(B)[/itex], and the proof is done.

I'll let you prove the union one, as you should know how to do it now. Although I'm not sure, my guess would be that g(A) U g(B) is a subset of g(A U B). Note how it's kind of like the "backwards" version of the previous proposition. Intersection and Union have that type of relationship, i.e. if you have a proposition with unions in it, then you can sort of "invert" things and change unions to intersections to get another true proposition.

For example, if we let A' denote the complement of A, then:

[tex]A\prime \cap B\prime = (A \cup B)\prime[/tex]

The intersection of complements is the complement of a union. If you take the complements first, then you take the intersection of the sets. If you "invert" the order, and take the complement last, then you union the sets first.
 
Oh and one more thing,

I know that (A n B) n C = A n (B n C)
and that A n (B u C) = (A n B) U (A n C).

I remember those properties from another class I took but I cannot remember what they are called. Can someone tell me this?
 
first one is associative, second is distributive (just like algebra),,, there are many more including DeMorgan's laws
 

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