Disproving A=B with Counter Example: Sets A, B & C

[klamgade]

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.

 

[gopher_p]

Are you convinced that the statement is false?

 

[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 $A=Y$.

 

[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?

 

[klamgade]

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

 

[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

 

[FactChecker]

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.

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.

 

[PeroK]

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.

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.

 

[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 $(a,c)\in A\times C$.

 

[PeroK]

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 $(a,c)\in A\times C$.

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:

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