Is the finite complement topology on R^2 the same as the product topology on R^2 that results from taking the product of two finite complement topology? why?
A subbasis element for the finite complement topology on $\displaystyle R^2$ has a form $\displaystyle R^2 - \{(a,b)\}$, $\displaystyle a,b \in R$.
A subbasis element for the product of two finite complement topology has a form $\displaystyle ((R -\{a\}) \times R) \cup (R \times (R - \{b\}))$, $\displaystyle a,b \in R$.
Both are equivalent.
I have only just returned from holiday, and my brain may be bleached out by the sun. But my thinking is that the set $\displaystyle (\mathbb{R} - {a})\times \mathbb{R}$ is open for the product of the 1-dimensional finite complement topologies, but not for the 2-dimensional finite complement topology (because its complement obviously contains infinitely many points).
Welcome back, Opalg !!
A projection map used in the above problem is open, but not continuous. So, I gave a subbasis element $\displaystyle ((R -\{a\}) \times R) \cup (R \times (R - \{b\}))$ rather than $\displaystyle ((R -\{a\}) \times R)$ or $\displaystyle (R \times (R - \{b\}))$ .
Although I was not able to prove that $\displaystyle ((R -\{a\}) \times R) \cup (R \times (R - \{b\}))$ is a subbasis element for the above problem, I did not find any counterexample or contradiction that it is not a subbasis element for the above problem either.
Please correct me if I am wrong.
I'm not sure what you mean by that. The product topology is defined to be the weakest topology on the product space that makes the projection maps (onto the coordinate spaces) continuous.
The inverse image of the set $\displaystyle \mathbb{R} -\{a\}$ under the projection map onto the first coordinate is $\displaystyle (\mathbb{R} -\{a\})\times\mathbb{R}$. So I claim that that set has to be open in the product topology.
You are definitely right.
Now, going back to the original question,
"Is the finite complement topology on $\displaystyle \mathbb{R}^2$ the same as the product topology on $\displaystyle \mathbb{R}^2$ that results from taking the product of two finite complement topology? why?"
Since the projection maps onto the coordinate spaces are not continuous for the above problem, the product topology on $\displaystyle \mathbb{R}^2$ by taking the product of two finite complement topology cannot even be defined in the first place.
Is that right?
No, that's not right. The product topology can always be defined.
In this example, the projection maps onto the coordinate spaces are not continuous when considered as maps from the space $\displaystyle \mathbb{R}^2$ with the finite complement topology to the space $\displaystyle \mathbb{R}$ with the finite complement topology. Therefore the finite complement topology on $\displaystyle \mathbb{R}^2$ is different from the product topology on $\displaystyle \mathbb{R}^2$ that arises from taking the finite complement topology on the two coordinate spaces. In fact, the product topology is stronger, since it contains more open sets.
I am having a hard time understanding what you said.
If I am able to resolve the below contradiction I've found, I might be able to grasp the whole idea.
"If a product topology T on $\displaystyle \mathbb{R}^2$ taking the product of two finite complement topology is defined, there exists a subbasis element for T that is not open".
According to the definition of a product topology on wiki, "...The product topology on X is defined to be the coarsest topology for which all the projections $\displaystyle p_i$ are continuous..".
If all the projections $\displaystyle p_i$ are continuous, a typical subbasis element defined by a product topology $\displaystyle p_i^{-1}(U)$ is open, and there is no contradiction.
To the best of my knowledge, there is no restriction that all the projections $\displaystyle p_i$ are continuous for a box topology.
Thus, we can define a box topology on $\displaystyle \mathbb{R}^2$ taking the product of two finite complement topology. I think it is not the case for the product topology.
Please correct me if the above argument is wrong.
Thanks for all your comments so far. It is a great learning experience for me.
Now I'm having a hard time understanding what you said! When you say that there is a subbasis element for T that is not open, what topology are you thinking of when you say "open"? The subbasis sets for the product topology are the sets of the form $\displaystyle p^{-1}(U)$, where U is an open set in one of the coordinate spaces, and p is the projection onto that coordinate. These sets are by definition open in T.
That is correct. And for a product of two spaces, the "box topology" is the same as the product topology. A basis for the product topology consists of all "boxes" of the form U×V, where U and V are open in the coordinate spaces.