# Sets and Logic

1. Sep 13, 2014

Hi ,

Can anyone please give me an idea to disprove the following with counter example:
A , B & C be sets. If A X C = B X C , the A = B .

I tried giving random numbers in venn diagram but didn't work. And, using subset way to prove equal but still couldn't solve it.

2. Sep 13, 2014

### gopher_p

Are you convinced that the statement is false?

3. Sep 14, 2014

### economicsnerd

Let $Y$ be some set (say $Y=\{0,1\}$ or $Y=\mathbb R$), and consider the special case where $C=Y^\infty$, the set of sequences in $Y$.

Does $A\times Y^\infty = B\times Y^\infty$ imply $A=B$? Let's think about the case of $A=Y$.

4. Sep 14, 2014

### caveman1917

Use the modus tollens. $A \times C = B \times C \Rightarrow A = B$ is the same as $A \neq B \Rightarrow A \times C \neq B \times C$. Can you find a counterexample to the latter statement?

Last edited: Sep 14, 2014
5. Sep 15, 2014

Thanks. I tried keeping set C as an empty set. It worked. Cheers :)

6. Oct 6, 2014

### SSGD

I'm new to Sets so if this doesn't apply my apolagizes.

AxC=BxC this statement will only be valid if C is invertable. So the Det(C) cannot be equal to zero. If you use a random number generator there is a possibility that the Det(C) will equal zero.

AxC=BxC
(AxC=BxC)xC^-1
A=B

if C is not invertable then A=B is not valid

7. Oct 8, 2014

### FactChecker

Your doubts are correct, but you still might be close.

In general, a set just sits there and does nothing. You can't talk about it being invertible. To talk about an inverse, there has to be more structure to C than just being a general set. Some sets are ordered pairs that represent functions. For a set like that, you can talk about the inverse of the function that the set defines.

Since the OP was asking for a counterexample, you could define a specific C that is a function and use that to prove it is a counterexample. You would have to be specific about C and use that. But the definition of the Cartesian product, A x C doesn't give you much to work with.

Consider the example that @klamgade came up with, C = $\emptyset$. It is the fact that the operation "x$\emptyset$" is not invertible that makes it work as a counterexample. It would not be correct to say that $\emptyset$ is not invertible.

8. Oct 8, 2014

### PeroK

The best way to show this is false, perhaps, is to prove it's true! As follows:

$Let \ \ c \in C$
$For \ \ a \in A$
$(a, c) \in A \times C$
$(a, c) \in B \times C$
$a \in B$
$\therefore \ A \subseteq B$

By the same method you can show $B \subseteq A$ hence $A = B$

So, the statement is true as long as C is non-empty.

9. Oct 8, 2014

### economicsnerd

I don't buy that argument. I don't think it follows from $(a,c)\in A\times C$ and $c\in C$ that $a\in A$. Let $C= \mathbb R^\infty$ and $A=\mathbb R$. Let $c=(1,1,1,...)\in C$ and $a = (0,0)\in \mathbb R^2.$ Then $a\notin A$, but $(a,c)=(0,d)$, where $d=(0,1,1,1,...)\in C$, so that $(a,c)\in A\times C$.

10. Oct 8, 2014

### PeroK

You're missing a subtlety of the definition of the Cartesian Product.

Let $A = \mathbb{R} \ \ and \ \ B = \mathbb{R}^2$

Then a typical element of $A \times B$ would be:

$(x, (y, z))$

Which is not the same as:

$((x, y), z) \ \ or \ \ (x, y, z)$

Which are elements of $B \times A$ and $\mathbb{R}^3$ respectively.

In any case:

$(a, b) \in A \times B \ \ iff \ \ a \in A \ \ and \ \ b \in B$