Prove that the topology of R^2 is the same as the topology of the the cartesian product RxR

Question

How to prove that the topology of R^2 is the same as the topology of the the cartesian product RxR? Is some bijective function needed?

Answer 1

I don't know what you mean by "the" topology of $\mathbb{R}^2$ or $\mathbb{R}\times\mathbb{R}$, but I assume you mean the open ball topology on $\mathbb{R}^2$ and the product topology of the open ball topology on $\mathbb{R}$ for $\mathbb{R}\times\mathbb{R}$.

In this case, it is sufficient to show that an open set in $\mathbb{R}^2$ is open in $\mathbb{R}\times\mathbb{R}$ and vice-versa.

If $U$ is open in $\mathbb{R}^2$, then for each $x\in U$ there is an open ball with radius (say) $\epsilon$ around $x$ which is contained in $U$. Then take the open box $(x-\epsilon/4,x+\epsilon/4)^2\in\mathbb{R}\times\mathbb{R}$. This open box is contained within the open ball (why?) and thus $U$ is open in $\mathbb{R}\times\mathbb{R}$.

You should be able to show the other direction from here.