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Set Theory

  • Thread starter trixitium
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  • #1
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Homework Statement



Show that:

[itex] A \cap B \subset (A \cap C) \cup (B \cap C') [/itex]

Homework Equations




The Attempt at a Solution



I tryed distribute [itex] (A \cap C) over (B \cap C') [/itex] but i'm always walking in circles and i don't came to a satisfactory answer. This exercise was in a section "some easy exercies on complementation" but i don't see how to use complements here.

Thanks
 

Answers and Replies

  • #2
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Is [itex]C^\prime[/itex] supposed to be the complement of C??
 
  • #3
HallsofIvy
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The standard way to show that "[itex]X\subseteq Y[/itex]" is to start "if [itex]x\in X[/itex]" and then use the definitions of X and Y to conclude "[itex]x\in Y[/itex]".

Here, if [itex]x\in A\cap B[/itex], what can you say about x?
 
  • #4
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Yes, [itex]C'[/itex] is the complement of [itex] C [/itex]

if [itex] x \in A \cap B [/itex],

by the definition of intersection:

[itex] A \cap B = \{x \in A : x \in B\} [/itex]

and we can conclude that x is simultaneously in A and B.

But my doubt, is how to reduced the [itex] (A \cap C) \cup (B \cap C') [/itex] to a expression that i can readly see that [itex] A \cap B \subset (A \cap C) \cup (B \cap C') [/itex].

I tryed ...

[itex] (A \cap C) \cup (B \cap C') = [/itex]
[itex] (A' \cup C')' \cup (B' \cup C)' = [/itex]
[itex] [(A' \cup C') \cap (B' \cup C)]' = [/itex]
[itex] (...) [/itex]
[itex] A \cup B \cup C' [/itex]

But it takes me a lot of work, i'm not sure if this result is correct and i think that exists a better way of doing this...

Thanks
 
  • #5
ehild
Homework Helper
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You certainly know that [itex]X\subset Y[/itex] if and only if [itex]X \bigcap Y=X[/itex].
[itex]X=A\bigcap B [/itex] and [itex]Y=(A \cap C) \cup (B \cap C')[/itex]

ehild
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
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Yes, [itex]C'[/itex] is the complement of [itex] C [/itex]

if [itex] x \in A \cap B [/itex],

by the definition of intersection:

[itex] A \cap B = \{x \in A : x \in B\} [/itex]

and we can conclude that x is simultaneously in A and B.

But my doubt, is how to reduced the [itex] (A \cap C) \cup (B \cap C') [/itex] to a expression that i can readly see that [itex] A \cap B \subset (A \cap C) \cup (B \cap C') [/itex].
Good, we know x is in both A and B. And we know that x is either in C or it is NOT! That means x is C or it is in C'
Case 1: Suppose x is in C. We know it is in A therefore ....
Case 2: Suppose x is in C'. We know it is in B therefore ....

I tryed ...

[itex] (A \cap C) \cup (B \cap C') = [/itex]
[itex] (A' \cup C')' \cup (B' \cup C)' = [/itex]
[itex] [(A' \cup C') \cap (B' \cup C)]' = [/itex]
[itex] (...) [/itex]
[itex] A \cup B \cup C' [/itex]

But it takes me a lot of work, i'm not sure if this result is correct and i think that exists a better way of doing this...

Thanks
In my opinion, to much focus on "formulas", not enough on basic "definitions".
 

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