If $G_1, G_2$ be two open sets in the usual metric space $(\mathbb{R}, d_1)$ can $G_1\times G_2$ be open in $\mathbb{R}^2$?

Question

Let us consider the metric spaces $\mathbb{R}$ and $\mathbb{R}^2$ equipped with usual metrics $d_1, d_2$ respectively.

If $G_1, G_2$ be two open sets in $\mathbb{R}$ then justify that $G_1\times G_2$ is an open set in the metric space $(\mathbb{R}^2, d_2)$.

I was studying metric spaces while this question came to my mind. I am not sure if my question is correct or not.

In case, if it is incorrect please help me by editing it and give me a correct version of it. Else please help me to prove it.

What I have tried so far : Let $(g_1, g_2)\in G_1\times G_2$. Since both $G_1, G_2$ are open sets so there exist open balls $B(g_1, r_1), B(g_2, r_2)$ for some $r_1>0, r_2>2$ such that \begin{align} &g_1\in B(g_1, r_1)\subseteq G_1\\ &g_2\in B(g_2, r_2)\subseteq G_2 \end{align} After this I dont know how to get an open ball for $B((g_1, g_2), r)$ for some positive $r$.

Is my approach correct ?

@mathcounterexamples.net That's not really relevant here - the question is not about the product topology, rather the topology on $\Bbb R^2$ induced by the standard metric. — David C. Ullrich, May 19 '18 at 16:25
@JoséCarlosSantos I have shown my attempt but not sure if it is correct — KON3, May 19 '18 at 16:28

José Carlos Santos · Accepted Answer · 2018-05-20T06:07:58.380

1

Take $r=\min\{r_1,r_2\}$. Then, if $(x_1,x_2)\in B\bigl((g_1,g_2),r\bigr)$, you know that$$\sqrt{(x_1-g_1)^2+(x_2-g_2)^2}<r.$$But then both numbers $|x_1-g_1|$ and $|x_2-g_2|$ are smaller than $r$ and$$|x_1-g_1|<r\implies|x_1-g_1|<r_1\implies x_1\in G_1$$and$$|x_2-g_2|<r\implies|x_2-g_2|<r_2\implies x_2\in G_2$$and therefore $(x_1,x_2)\in G_1\times G_2$. This proves that$$B\bigl((g_1,g_2),r\bigr)\subset G_1\times G_2.$$

edited May 20 '18 at 06:07

answered May 19 '18 at 16:39

José Carlos Santos

427,504

Its really beautiful. I have a doubt only. The definition of usual metric $d_1$ in $\mathbb{R}$ is not same as usual metric $d_2$ in $\mathbb{R}^2$, isn't it ? So are we allowed to construct $D$ like this ? – KON3 May 20 '18 at 02:29
@Anjan3 I meant $B$, not $D$. I've edited my answer. I hope that everything is clear now. – José Carlos Santos May 20 '18 at 06:08
Ain't working with inequalities grand?LOL – Mathemagician1234 May 20 '18 at 06:09

If $G_1, G_2$ be two open sets in the usual metric space $(\mathbb{R}, d_1)$ can $G_1\times G_2$ be open in $\mathbb{R}^2$?

1 Answers1