# Thread: Define function for any set of real numbers..

1. ## Define function for any set of real numbers..

Hi,
first of all, Merry Christmas! I am doing some math while digesting another huge dinner, and have a small question.

Problem:
If A is any set of real numbers, define a function $C_A$ as follows:
$C_{A}(x)=\left\{ \begin{array}{cc}
1, & \mbox{x in A}\\
0, & \mbox{x not in A}\\
\end{array} \right.
$

Find expressions for $C_{A\cap B}$ and $C_{A\cup B}$ and $C_{\Re-A}$

Attempt:
The way I am reading this, is that I first have to define a function which is 1 for real numbers, and 0 for pure imaginary numbers.
I think this one works fine:

$C_{A}(x)=\frac{x+\overline{x}}{2x}$ where $\overline{x}$ is the complex conjugate.

I am not so sure about the other expressions. If I still take B to be the set of pure imaginary numbers am I suppose to define the following functions:

$C_{A\cap B}(x) = \left\{ \begin{array}{cc}
0, & x\in A \\
0, & x\in B \\
\end{array}\right.$

$C_{A\cup B}(x) = \left\{ \begin{array}{cc}
1, & x\in A\\
1, & x\in B\\
\end{array}\right.$

$C_{\Re-A}(x) = \left\{ \begin{array}{cc}
0, & x\in A\\
0, & x\in \Re \\
1, & x\in \Re \quad x \notin A\\
\end{array}\right.$

2. Originally Posted by Mollier
Problem:
If A is any set of real numbers, define a function $C_A$ as follows:
$C_{A}(x)=\left\{ \begin{array}{cc}
1, & \mbox{x in A}\\
0, & \mbox{x not in A}\\
\end{array} \right.
$

Find expressions for $C_{A\cap B}$ and $C_{A\cup B}$ and $C_{\Re-A}$
Why would you introduce complex numbers? These are sets of real numbers.
Here is the first, $C_{A\cap B}(x)=C_{A}(x)\cdot C_{ B}(x)$.
Now why is that true?

You try the next two.

3. Originally Posted by Plato
Why would you introduce complex numbers?
Because I am a fool.

Originally Posted by Plato
Here is the first, $C_{A\cap B}(x)=C_{A}(x)\cdot C_{ B}(x)$.
Now why is that true?
Well, If x is in A and B, then $C_A=1$ and we are left with $C_B$
If x is in A, but not in B then $C_B(x)$ is not defined.
If x is not in A, $C_{A\cap B}=0$.

Originally Posted by Plato
You try the next two.
$C_{A\cup B}(x) = C_A(x)+C_B(x)$
The functions domain is x in A or B, so it is not defined if x is not in A or B..

$C_{\Re-A}(x)=(C_A(x)-1)C_B(x)$
If x is in A then $C_A(x)=1$ and $C_{\Re-A}=0$

I'm not very good at this...

4. Originally Posted by Mollier
$C_{A\cup B}(x) = C_A(x)+C_B(x)$
The functions domain is x in A or B, so it is not defined if x is not in A or B..

$C_{\Re-A}(x)=(C_A(x)-1)C_B(x)$
If x is in A then $C_A(x)=1$ and $C_{\Re-A}=0$
Not quite.
Try, $C_{A\cup B}(x) = C_A(x)+C_B(x)-C_{A\cap B}(x)$
And, $C_{\Re-A}(x)=1-C_A(x)$

Now you reply with reasons why.

5. Say A = 1,2,3,4,5 and B=3,4,5,6,7 then
$A\cup B$= 1,2,3,4,5,6,7.
We want to remove the overlapping entries.
It makes sense if I think of $C_{A\cup B}$ as the probability of picking an entry from A or B..

$C_{\Re-A}(x)=1-C_A(x)$
is zero when $x\in A$ and 1 otherwise.

Worst thing is, I was quite good at this a million years ago

Thank you very much.