About problem of baby Rudin basic topology

Question

In the charter of basic topology Rudin says (a,b) is no open if we regard it as a subset of $R^2$,

I don’t understand why ,is he mean the region x$\in (a,b),y\in (-\infty ,+\infty )$

I think every point of the region is interior point

Answer 1

The set $(a,b)$ can be looked at as subset of the real numbers, thus $(a,b) = \{x \in \mathbb{R}: a < x < b\}$ and since (using Rudin's notation of neighborhoods)

$$(a,b) = N_{\frac{b-a}{2}}\left(\frac{a+b}{2}\right)$$

the set is open as viewed as a subset of $\mathbb{R}$ (Rudin already proved that neighborhoods are open).

However, we can look at $(a,b) \subseteq \mathbb{R} \subseteq \mathbb{C}$ and then the situation changes. We view this segment in the complex plane, and not one point is interior point in the complex plane. This is geometrically clear: take any neighborhood of a point on the segment, and it will have points that are not on the segment.

Note that Rudin identifies $\mathbb{R}^ 2$ and $\mathbb{C}$.

Other note: The answer by BigbearZzz is correct as well. Through the natural identification $a+bi \mapsto (a,b)$ the set I wrote down corresponds to the set $\{(x,y) \in \mathbb{R}^ 2\mid a < x < b, y = 0\}$.

Answer 2

I believe he meant the set $$ (a,b) = \{ (x,y)\in\Bbb R^2 : x\in(a,b)\text{ and } y=0 \} $$ which is a subset of $\Bbb R\times \{0\}.$