Fill in the blank to make true identities: $(A\Delta B)\cap C=(C\setminus A)\Delta(\_\_\_)$, $A,B,C$ set of real numbers.

Question

I'm come across a problem regarding set theory and operations, it goes like this:

Fill in the blank to make true identities $$(A\Delta B)\cap C=(C\setminus A)\Delta(\_\_\_)$$

Where $A$, $B$ and $C$ are set of real numbers.

The problem that I encounter is when I fill out $(A\Delta B)\cap C$, I got this as the following:

Figure 1. $(A\Delta B)\cap C$

and when I fill out $(C\setminus A)$ the graph is going to look like this: graph of (C\A)

but $(\Delta)$ denoted as the symmetric difference between the two set, or the shaded region will be in the sets but not at their intersection. However, given the condition in which I have to make the identity true, $(\Delta)$ is not sufficient to cancel out the elements of C to make the right side of the equation look like the left side of the equation. Could anyone give me a hint on what to do? Thank you

Answer 1

As I see, you already found out in your diagram, what the left site of your equation is.

Now on the right site of your equation, you have $(C \backslash A)\Delta X$. $X$ is then the union of what you want to "add" to $(C \backslash A)\Delta X$ and of what you want to remove from $(C \backslash A)\Delta X$, so that it gives you the left site. This is just because of the definition of $\Delta$. Make it clear to you with an easy example.

By using your diagram, you can see then that $X = C \backslash B$.

Answer 2

How about something like this: $$ blank = C- B $$

Answer 3

Hint

If $(A\Delta B)\cap C = (C\setminus A)\Delta(\underline\quad)$

Then by symmetry: $(B\Delta A)\cap C = (C\setminus B)\Delta(\underline\quad)$

(Just verify that there is no third term and your should be done.)