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".