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Intro

Hello, I'm starting to read Munkres' Topology, and I'm having a bit of trouble solving problems involving the cartesian product. For example, there's this problem:

Sample Exercise

$$\textit{Is the following true?}$$ $$(A-B)\times(C-D)=(A\times C-B\times C)-A\times D$$

Erroneous Solution

I have a really hard time visualizing this kind of thing, so I tried to sort of give a half-explanation: Since it is true that

$$(A\times C-B\times C)=(A-B)\times C$$

We're left with

$$(A-B)\times(C-D)=(A-B)\times C-A\times D$$

Now, what we have on the right-hand-side of the original equation is very similar to the left-hand-side except for one thing: for the $(-A\times D)$ part, we should have $(-(A-B)\times D)$ if we wanted both sides to be the same. It's pretty easy to convince yourself that: $(A\times D)\supseteq(A-B)\times D) $. Therefore, we must have that

$$(A-B)\times(C-D)\supseteq(A\times C-B\times C)-A\times D$$

Which means that the initial statement is not true.

The Problem

My answer was obviously incorrect; I looked online for a solution set and the correct answer was that the given statement is correct. Where did I go wrong here? Also- is there an 'official' way way of approaching these types of exercises? For exercises involving concepts like union and set difference I've had no problem. This is because I find unions and set differences very easy to conceptualize, or at least draw in a Venn diagram. Is there a similar diagram I could draw for cartesian products which would be useful? Or a more systematic way of approaching these exercises?

Thanks in advance.

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Don't visualize. Use the definitions.

Visualization can be helpful, but it can be misleading, and if you don't learn to work out the definitions, you can get stuck quite easily because you spend all your energy trying to visualize the problem.

In one direction, if $(a,c)\in(A-B)\times (C-D)$ then $a\in A, c\in C$ and $a\notin B,c\notin D$, therefore $(a,c)\notin B\times C$ as well $(a,c)\notin A\times D$. Therefore $(a,c)\in (A\times C-B\times C)-A\times D$.

The other direction is as simple.


One useful tip is to learn how to simplify statements like the one on the RHS. Relevant to this question is the identity, $(A-B)-C=A-(B\cup C)$.

This makes it easier to see why equality holds: If $(a,c)$ is in the LHS, then $(a,c)\in A\times C$ and $(a,c)\notin B\times C$ as well $(a,c)\notin A\times D$. Therefore $\subseteq$ holds. And so on.

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Check both inclusions, this is how this is done always. I would not say that your proof is false from the beginning, but its just incomplete. $A\subset B$ does not exclude $A=B$. You say: Therefore, we must have that

$$(A-B)\times(C-D)\supseteq(A\times C-B\times C)-A\times D$$

Which means that the initial statement is not true. How does this make sense? Two sets are equal iff $A\subset B$ and $B\subset A$! Just check the other inclusion.

Cartesian products $X\times Y$ can be drawn like $\mathbf{R}^2$: put $X$ on the $x$-axis, and $Y$ on the $y$-axis. (If you do it like this the result is actually obvious!)

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  • $\begingroup$ Yeah, that's correct. I completely messed up on the definition of $A=B$. $\endgroup$ – Physics Llama Jul 11 '14 at 21:56

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