Question

is (a b) open on r2?

Answer 1

you asked the same question yesterday...I gave you a proof...are you not satisfied? I can give you a different proof....

Answer 2

Supposed \(a,b\in \mathbb{R}\) where \(a<b\).
Choose any \(x\in (a,b)\) and consider \(\epsilon = \min(|x-a|,|x-b|)=\min(x-a,b-x)\).
It is obvious that \(B(x,\epsilon)\subset (a,b).\) Showing \((a,b)\) is an open set in \(\mathbb{R}\). \(_\square\)

Answer 3

A set \(S\) is open if
\(\forall \ x\in \ S, \ \ \ \exists \ \epsilon >0 \ \ \ \ s.t. \ \ \ B(x,\epsilon)\subset S\).

Answer 4

Also the way you asked the question does not make sense

\((a,b)\subset \mathbb{R}\) not \(\mathbb{R}^2\)