I know each open set not necessarily be the cartesian product of two open sets. but I want an example for this equation. Please help me. Thank you in advance!
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Sorry. What is an example of an open set A which we can write it in the form of a Cartesian product of two sets B and C? (B and C not necessarily be open) – Imaney Apr 15 '21 at 12:30
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Take one of the sets to be empty. – Kavi Rama Murthy Apr 15 '21 at 12:33
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1If you start with open sets $B,C$ then $B\times C$ will be open (in the appropriate topology). Unless $B,C$ are both open, their product can't be open. – Gerry Myerson Apr 15 '21 at 12:40
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I think empty is an open set. according to this link. – Imaney Apr 15 '21 at 12:44
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To be honest, I wanna prove "Each open set is not a Cartesian product of two open sets.". How can I prove it? @GerryMyerson – Imaney Apr 15 '21 at 12:49
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To begin with, if $B$ is open in the topological space $X$, and $C$ is open in the topological space $Y$, then you have to tell us how you are defining a topology on $X\times Y$, in terms of the topologies on $X$ and on $Y$. – Gerry Myerson Apr 15 '21 at 12:53
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So, what's your definition, Iman? (lman?) – Gerry Myerson Apr 17 '21 at 03:18
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1Actually Iman (with i). It's an Arabic name. It means believing in something like a god or a person. @GerryMyerson – Imaney Apr 17 '21 at 08:06
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$$B = (0,1)$$ $$C = (0,1)$$ $$A = B \times C = (0,1) \times (0,1)$$
It is (hopefully) clear that $B$ and $C$ are both open sets, since they are open intervals. To see why $A$ is open, show that every point in $A$ is an interior point.
Hint: Assume that arbitrary point in $A$ is $p = (x,y)$ where both $x$ and $y$ are in $(0,1)$. Now, try to find a ball around $p$ such that the whole ball is contained in $A$.
To answer your question in comments (whether every open set is cartesian product of two open sets):
No. For a simple counterexample, consider a open disk in $\mathbb{R}^2$. Do you see why this set cannot be Cartesian product of two open intervals in $\mathbb{R}$?
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