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Consider $\mathbb{R}^2$ with the included point topology (open subsets are those containing $(0,0)$); and $\mathbb{R}$ with the same topology (open subsets are those containing $0$). Now consider the product topology $\mathbb{R}\times \mathbb{R}$ with each one having the topology explained before.Can $\mathbb{R}×\mathbb{R}$ and $\mathbb{R}^2$ be homeomorphic with this topology?

I can only think about trying to see if it is possible to have a continuous bijection between this two sets and see if it can be an open map, but I'm stuck on proving this, any help would be appreciated, thanks!

L. Sandoval
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  • Consider the bijection presented in this question. Note that in this bijection, $(0,0)\mapsto0$ which means that the bijection is a homeomorphism. – Rushabh Mehta Nov 15 '19 at 21:22
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    $\mathbb{R}^2$ with the topology of sets that contain $(0,0)$ has ${(0,0),(a,b)}$, with $(a,b)\neq(0,0)$, of cardinality $2$, as some of the open sets. However, $\mathbb{R}\times \mathbb{R}$ with the product topology coming from the factors having the topology of the sets that contain $0$, contains only sets of cardinality $2$ of the form ${(0,0), (0,a)}$ or ${(0,0), (a,0)}$. The only open singleton in both topologies is ${(0,0)}$. So, one needs to send $(0,0)$ to $(0,0)$. But all other points would have to be send to the axes of $\mathbb{R}\times\mathbb{R}$. – conditionalMethod Nov 15 '19 at 21:36
  • I don't fully understand why does f(0,0)=(0,0), and why does that imply all other points would have to be sent to other than (0,0) ? is it because its a bijection? – L. Sandoval Nov 15 '19 at 22:09
  • The set ${(0,0)}$ is open in both spaces and it is the only open singleton in both. This implies that $f(0,0)=f(0,0)$. Since $f^{-1}({(0,0)})$ would have to be both open and a singleton. – conditionalMethod Nov 15 '19 at 22:14
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    Assume that $f:\mathbb{R}^2\to\mathbb{R}\times\mathbb{R}$ is open and bijective. Then it would have to send the open set ${(0,0),(a,b)}$, with $(0,0)\neq(a,b)$, to some open set of cardinality $2$. The only open sets of cardinality $2$ in $\mathbb{R}\times\mathbb{R}$ are of the form ${0}\times {0,c}$ or ${0,c}\times{0}$. So, $(a,b)$ will have to be send to some point of the form $(c,0)$ or of the form $(0,c)$. – conditionalMethod Nov 15 '19 at 22:16
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    Why do you write $\Bbb R\times\Bbb R$ and $\Bbb R^2$, don't you mean $\Bbb R\times\Bbb R$ and $\Bbb R$ ? Are you considering $\Bbb R^2$ with two different topologies (I think no), or are you considering one topology on $\Bbb R\times\Bbb R$ and another topology on $\Bbb R$ (seems likely), but then why would you write "included point topology in $\Bbb R\times\Bbb R$ and in $\Bbb R^2$? What is the difference between $\Bbb R\times\Bbb R$ and $\Bbb R^2$? – Mirko Nov 15 '19 at 23:21
  • I agree with @Mirko - the comparison here seems to be between $\mathbb R\times\mathbb R$ and $\mathbb R$. – Math1000 Nov 16 '19 at 00:49
  • @Mirko We are considering $\mathbb{R}^2$ with the topology given by the included point topology ($U$ is open if (0,0) is in $U$), and then considering $\mathbb{R}$ again with the topology given by the included point topology ($U$ open if $0\in U$). But the homeomorphism we want to see is between the product topology $\mathbb{R}\times\mathbb{R}$ (each one with the topology explained before) and $\mathbb{R}^2$ with the included point topology explained before. – L. Sandoval Nov 16 '19 at 10:26
  • @L.Sandoval Could you please edit your question to reflect the clarification from your previous comment? – Mirko Nov 16 '19 at 15:22
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    Done, edited the question to clarify. Now I understood the answer, you can not find any homeomorphism because (with the facts explained in the previous comments) $f(0,0)=(0,0)$ and $f(a,b)=(0,c)$ or $(c,0)$, so it can not be a bijection because for example $(1,2)$ would not have pre-image, so it can't be surjective. – L. Sandoval Nov 16 '19 at 16:23
  • @conditionalMethod I think your comments are correct, why don't you post them as an answer? L. Sandoval, yes (assuming $f:\Bbb R^2\to\Bbb R\times\Bbb R$). Alternatively: Every two-point set $D$ containing $(0,0)$ is open in $\Bbb R^2$, but some two-point sets containing $(0,0)$, like ${(0,0),(1,2)}$, are not open in $\Bbb R\times\Bbb R$. (And $(0,0)$ is the only point that is open in both topologies.) – Mirko Nov 16 '19 at 17:02

2 Answers2

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A way to understand when a topology $\cal T$ in a product $X=X_1\times X_2$ is a product topology ${\cal T}_1\times{\cal T}_2$ is that we know the candidates to factor topologies: the projection $X\to X_1$ induces homeos $\{y=y_0\}\to X_1$, and similarly for the other projection. In our case, ${\cal T}|\{y=1\}$ is discrete and so ${\cal T}_1$ cannot be a point topology.

Jesus RS
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The open sets of R$^2$ are { U : (0,0) in U }.

Each open set of R×R is a union of sets of the form U×V where U,V are open with the special point 0 topology of R.
Thus (0,0) is in every open set of R×R.
Furthermore if (0,0) is in U then U is the union of sets for the form
{0}×{0,r}, {0,r}×{0}, r in R. Thus U is open within R×R.

Consequently, as R$^2$ and R×R have identical open sets, the identity map a homeomorphism.

William Elliot
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  • If U contains just two points, (0,0) and (1,1) then U is not the union of sets for the form {0}×{0,r}, {0,r}×{0}, r in R. I would think that the comments by @conditionalMethod provide a negative answer. – Mirko Nov 16 '19 at 16:09
  • Some of your U are open in R$^2$, some (from U×V) open in $R$. It is difficult to follow which is which somewhere around the middle of your answer. Perhaps involve W instead of one of those U? – Mirko Nov 16 '19 at 16:20
  • The open sets of $\mathbb{R}\times\mathbb{R}$ are not all unions of sets of the form ${0}\times{0,r}$ and ${0,r}\times{0}$. For example ${0,1}\times{0,1}$ is open and contains the point $(1,1)$, which is not in the axes. The problem with $\mathbb{R}\times\mathbb{R}$ is that when one of its open sets contains point out of the axes, then the set also contain the projections of that point to the axes. This makes these types of open sets always have cardinality larger than $2$. – conditionalMethod Nov 16 '19 at 23:47
  • The impossibility of a homeomorphism with $\mathbb{R}^2$ (with the topology of sets containing $(0,0)$) follows by noting that in it there are open sets of cardinality $2$ that do contain points outside of the axes, like ${(0,0), (1,1)}$. They force all points of $\mathbb{R}^2$ to have to be sent to points on the axes of $\mathbb{R}\times\mathbb{R}$ for the map to be open. But then it cannot be onto. – conditionalMethod Nov 16 '19 at 23:48