Homeomorphism between $(\Bbb R \times \Bbb R, \tau_{\Bbb R \times \Bbb R})$ and $(\Bbb R^2,\tau_E)$

Question

Consider $(\Bbb R, \tau_E)$, i.e. $\Bbb R$ with Euclidean topology and the product $\Bbb R \times \Bbb R$ with the product topology. Show that the resulting space is homeomorphic to $\Bbb R^2$ with its Euclidean topology.

I wanted to that in a rigorous way that means finding a bijective continuous map $f$ between $(\Bbb R \times \Bbb R, \tau_{\Bbb R \times \Bbb R})$ and $(\Bbb R^2,\tau_E)$ with bijective inverse.

This map $f$ must satisfy the fact that an open set in $(\Bbb R^2,\tau_E)$ has a unique open pre-image in $(\Bbb R \times \Bbb R, \tau_{\Bbb R \times \Bbb R})$. I think that we are looking for a map that sends bijectively a square to a circle. However, a circle in $\Bbb R^2$ is completely determinate by $3$ parameters : $2$ coordinates $(x,y)$ for its center and 1 parameter for the radius. Meanwhile an open set in $(\Bbb R \times \Bbb R, \tau_{\Bbb R \times \Bbb R})$ is a rectangle that is given by $4$ parameters : 2 coordinates for a point and 2 other parameters for length and width. I cannot see how the map can be injective in this case. There is some kind of "dimension" problem here...

Answer 1

You have $\tau_{\Bbb R \times \Bbb R} = \tau_E$.

It seems that you have doubts because a basis $B_E$ of $\tau_E$ is given by the open disks $D_r(x,y) = \{ (a,b) \in \mathbb R^2 \mid \lVert (a,b) - (x,y) \rVert < r \}$ whereas a basis $B_{\Bbb R \times \Bbb R}$ of $\tau_{\Bbb R \times \Bbb R}$ is given by the open rectangles $R_{(l,w)}(x,y) = \{ ((a,b) \in \mathbb R^2 \mid \lvert a - x \rvert < l, \lvert b - y \rvert < w \}$. At first glance there seem to be "more" rectangles than disks due to the number of parameters. However, you should be aware that we work with infinite sets: In fact $B_E$ and $B_{\Bbb R \times \Bbb R}$ have the same cardinality. So the number of parameters is irrelevant. But even if the cardinality would be different, we cannot conclude that the topologies generated by these bases are different. It is very well possible that a basis is unnecessarily big, i.e. contains dispensible elements. As an example take the space $X = \{0,1\}$ with the discrete topology. It has the two bases $\{\{0\},\{1\}\}$ and $\{\{0\},\{1\}, X\}$ with different cardinality.

To prove that $\tau_{\Bbb R \times \Bbb R} = \tau_E$ you have to show that

1. for each disk $D_r(x,y)$ and each $(a,b) \in D_r(x,y)$ there exists a rectangle $R_{(l,w)}(a,b) \subset D_r(x,y)$.

2. for each rectangle $R_{(l,w)}(x,y)$ and each $(a,b) \in R_{(l,w)}(x,y)$ there exists a disk $D_r(a,b) \subset R_{(l,w)}(x,y)$.

This is an easy exercise.