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Munkres topology book set theory chapter 1 question 10c

Let $\mathbb R$ denote the set of real numbers. for each of the following subsets of $\mathbb R\times\mathbb R$ determine whether it is equal to Cartesian product of two subsets of $\mathbb R$.

Now here $\{(x,y)| y\gt x\}$ is given as not equal to Cartesian product of two subsets of $\mathbb R$ but we can construct a product like that below:

$\{(x,y)| x \in \mathbb R, y\in B\text{ such that } B=\{y| y\gt x\}\}$

Still $x$ and $y$ are subsets of $\mathbb R$

Can anyone explain me why its not a true Cartesian product definition

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I think you meant to say "Still $\mathbf{R}$ and $B$ are subsets of $\mathbf{R}$." As stated in Matt Pressland's answer, the problem is that $B$ cannot be defined until after you've selected the element from $\mathbf{R}$. See my answer below for an argument showing that there is no way to choose sets $A$ and $B$ in advance so that $A\times B$ equals your set. – Will Orrick May 9 '12 at 9:44
up vote 4 down vote accepted

If $(a_1,b_1),(a_2,b_2)\in A\times B$ then $a_1,a_2\in A$ and $b_1,b_2\in B$. But then $(a_1,b_2)$ and $(a_2,b_1)$ are also in $A\times B$ since the latter is the set of all ordered pairs such that $a\in A$ and $b\in B$. Apply this to the set $S=\{(x,y)\mid y\gt x\}$. The pairs $(1,2)$ and $(3,4)$ are in $S$. If $S$ is a Cartesian product then $(3,2)$ must be in $S$. But this contradicts the definition of $S$.

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The set $B$ you have defined depends on $x$. For it to be a Cartesian product, you need to have elements $(x,y)$ such that $x\in A$ and $y\in B$, where $A$ and $B$ are fixed sets, not depending on the choice of either co-ordinate.

It is true that the set $\{(x,y):y>x\}$ is the union of products $\{x\}\times B_x$, where $B_x=\{y:y>x\}$, which is what you have said, but this isn't enough to make it a product in its own right.

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