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So I'm trying to understand my teachers solution to this question. She writes: Note that $$(\Bbb R\setminus \{0\})\times\Bbb R= \Bbb R^2\setminus (\{0\} \times\Bbb R)\subset\Bbb R^2$$ is open in the product topology and not in the cofinite topology (I see this). Every open set in the cofinite topology on $\Bbb R^2$ is a finite intersection of complements of singletons. (Im still with her here). For every $(x,y)\in\Bbb R^2$, the set $$\Bbb R^2\setminus \{(x,y)\}=((\Bbb R\setminus \{x\} \times \Bbb R) \cup (\Bbb R \times (\Bbb R\setminus \{y\}))$$ is open in the product topology. Hence the product topology is a strict refinement of the cofinite topology.

So its that last step I dont get. Is it just because that union is not open in the cofinite topology, it has to be finer then the product t since it then cant be in the cofinite t?

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By writing $$ \Bbb R^2\setminus \{(x,y)\}=((\Bbb R\setminus \{x\} \times \Bbb R) \cup (\Bbb R \times (\Bbb R\setminus \{y\})), $$ you've shown that the complement of any singleton is open in the product topology. Any set which is open in the cofinite topology is a finite intersection of complements of singletons, so it's a finite intersection of sets which are open in the product topology; that means it's open in the product topology.

By definition, if every set which is open in the cofinite topology is also open in the product topology, then the product topology is finer than the cofinite topology. (It's strictly finer than the cofinite topology since you can easily come up with sets, such as $\Bbb R \times (0, \infty)$, which are open in the product topology but not in the cofinite topology.)

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  • $\begingroup$ AAAaahaaa… thank you so much! :D $\endgroup$ – Linelina Dec 9 '19 at 15:50

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