If $C$ is a fixed non-empty set, which one of the following options imply $A=B$?
1) $A - C = B - C$
2) $A \cup C = B \cup C$
3) $A \cap C = B \cap C$
4) $A \Delta C = B \Delta C$
I'm not sure why all of them are not valid option.
The problem I'm having is with premise. If $A=B$ then it would imply that all of the options are true. Can you maybe expand that a bit more on how to reason the problem?
On
Your mistake is having the implication backwards. You are correct that $A=B$ implies each of them, but that isn't what is being asked. You are being asked which of them implies $A=B$. For (2), for example, there could be an element that is in $A$ and $C$, so is in both unions, but is not in $B$. That makes $A \neq B$ even though $A \cup C = B \cup C$
There are $2^3=8$ possibilities for an element to be in some combination of $A,B,C$. Because of the symmetry of $A$ and $B$ you can ignore the ones where it is in $B$ and not $A$ because the result will be the same for one that is in $A$ but not $B$. You can also ignore the ones where it is in both or neither of $A,B$. Consider an element that is in $A$, not in $B$, and maybe in $C$, so there are only two real possibilities. See if either of them work with your equalities. If either one does, that equality does not imply $A=B$.
On
The problem I'm having is with premise. If A=B then it would imply that all of the options are true.
Which is the exact reverse of what is being asked.
The question isn't what can we conclude from $A=B$. It is what can we use to conclude $A=B$.
If the question was "Which one of these imply Charlie is a mammal" and the first option was "Charlie breaths" then
Although "Charlie is a mammal" $\implies$ "Charlie breathes",
"Charlie breathes" $\not \implies$ "Charlie is a mammal".
So that is not an answer.
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1) A−C=B−C
The fails as $A$ may have different elements of $C$ than $B$ does.
2) A∪C=B∪C
Ditto.
3) A∩C=B∩C
This fails because $A$ may have different elements that are not is $C$ than $B$ does.
4) AΔC=BΔC
If believe $A \Delta C = (A\setminus C)\cup (C\setminus A)$. Am I correct?
If so $A\Delta C$ are all the elements of $A$ that are not in $C$ and all the elements in $C$ that are not in $A$. These are equal to all the elements of $A$ that are not in $B$ and all the elements in $C$ that are not in $B$.
So all the elements of $A$ that are not in $C$ are the same as the elements in $B$ that are not in $C$. And the elements in $C$ that are not in $A$ are the same of the elements of $C$ that are not in $B$.
That means the rest of the elements in $C$ that aren't in $A \Delta C$ are elements in $A$. And the elements in $C$ that aren't in $B \Delta C$ are elements in $B$. And this are the same elements.
So the elements of $A$ that are not in $C$ are the same as the elements of $B$ that are not in $C$. And the elements of $A$ that are in $C$ are the same as the elements of $B$ that are in $C$.
So $A$ and $B$ have the same elements.
This is the only one that applies.
You seem to have misunderstood what the problem is. Taking part (b) as an example, you are right that $$A=B \quad\text{ implies }\quad A\cup C = B\cup C $$ but that's not what is being asked here. Instead you're asked to determine whether $$ A\cup C = B\cup C \quad\text{ implies }\quad A=B $$
In this case the answer is that it doesn't, as shown by the counterexample $$ A = \varnothing, \quad B=\{1\}, \quad C=\{1\} $$ where $A\cup C=B\cup C$ is indeed true but $A=B$ isn't.