MHB Prove relationship between sets

[biocamme]

For any two sets A and B prove:

(A∪B)^c=A^c∩B^c
(A∩B)^c=A^c∪B^c

 

[Greg]

Hello biocamme and welcome to MHB! :D

I've edited the title of your thread to be more descriptive of the problem at hand. Also, we ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

A brief description of the notation you are using may be helpful to some.

 

[Monoxdifly]

For any two sets A and B prove:

(A∪B)^c=A^c∩B^c
(A∩B)^c=A^c∪B^c

By using Venn diagram?

 

[soroban]

\text{For any two sets }A\text{ and }B.\:\text{ prove: }\; \begin{array}{cc} (A \cup B)^c\:=\:A^c \cap B^c \\ (A \cap B)^c \:=\:A^c \cup B^c \end{array}

By Venn diagrams? . Truth tables? . Other?

 

[HallsofIvy]

A standard method for proving two sets, X and Y, equal is to prove first that X\subseteq Y and then that Y\subseteq X. And to prove X\subseteq Y, start "if x\in X" and then use the properties of X and Y to conclude x\in Y.

Here, if x\in (A\cup B)^c x is not in A\cup B. So x is not in A and x is not in B. Since x is not in A then it is in A^c . Since x is not in B, then it is in B^c so x is in A^c\cap B^c

Now, do the other way- if x is in A^c\cap B^c then it is in both A^c and B^c so x is not in A and not in B. That is, x is not in A\cup B so is in (A\cup B)^c.