Demonstrating Subset Relationship: A ∩ B ⊆ (A ∩ C) ∪ (B ∩ C')

[trixitium]

Homework Statement

Show that:

A \cap B \subset (A \cap C) \cup (B \cap C')

Homework Equations

The Attempt at a Solution

I tryed distribute (A \cap C) over (B \cap C') 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

 

[micromass]

Is C^\prime supposed to be the complement of C??

 

[HallsofIvy]

The standard way to show that "X\subseteq Y" is to start "if x\in X" and then use the definitions of X and Y to conclude "x\in Y".

Here, if x\in A\cap B, what can you say about x?

 

[trixitium]

Yes, C' is the complement of C

if x \in A \cap B,

by the definition of intersection:

A \cap B = \{x \in A : x \in B\}

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

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

I tryed ...

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

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

 

[ehild]

You certainly know that X\subset Y if and only if X \bigcap Y=X.
X=A\bigcap B and Y=(A \cap C) \cup (B \cap C')

ehild

 

[HallsofIvy]

Yes, C' is the complement of C

if x \in A \cap B,

by the definition of intersection:

A \cap B = \{x \in A : x \in B\}

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

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

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 ...

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

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".