Open sets in $\mathbb{R}$ In Apostol's Analysis, he says that an open set in $\mathbb{R}^1$ is no longer an open set if it is considered a subset of $\mathbb{R}^2$ because such a set cannot contain a 2-ball. I am having trouble understanding how an open interval in $\mathbb{R}^1$ can ever be a subset of $\mathbb{R}^2$.  Also, does the same apply for open sets in $\mathbb{R}^n$ as subsets of $\mathbb{R}^{n+1}$?   
 A: You are right to be suspicious. I think what is meant by: "considered as a subset of $\mathbb{R}^2$" is that if $S \subseteq \mathbb{R}$, the set: $S \times \left\{0 \right\} \subseteq \mathbb{R}^2$. It is then clear that such a set cannot contain an open ball and hence is not open, so long as $S \neq \emptyset$. 
And yes, the claim can be extended to general $n$.
A: the open Interval $(a,b)$ can be see in $\mathbb{R}^2$ as
$$\{(x,0)\hspace{0.2cm} | \hspace{0.2cm}x \in (a,b) \}$$
A: usually, we think of $\mathbb{R}^n$ as imbeded in $\mathbb{R}^{n+m}$ by $$(x_1,...,x_n)\rightarrow (x_1,...,x_n,0,...,0)$$(for this matter, any continuious embedding will do).
the claim is true for any $n,m>0$ ,because any ball in $\mathbb{R}^{n+m}$ has pointsd differing in all cordinates.
A: the open Interval $(a,b)$ can be see in $\mathbb{R}^2$ as
$$I_{ab}:=\{(x,0)\hspace{0.2cm} | \hspace{0.2cm}x \in (a,b) \}.$$
Suposed that $I_{ab}$ is a open set in $\mathbb{R}^2$, as $I_{ab}\neq \emptyset$, there's a ball of ratius $\delta>0$ centered in $(c,0)$ for some $c\in(a,b)$ (we will denote such  ball as $B_c(\delta)$) such that 
$$B_c(\delta)\subset I_{ab}.$$
Notice now that $(c,\frac{\delta}{2})$ belongs to $B_c(\delta)$ but $(c,\frac{\delta}{2})\notin I_{a,b}$ (--Why?). Absurd!
Therefore, none interval of $\mathbb{R}$ can be a open set in $\mathbb R^2$.
$\hspace{17cm}$$\Box$
Ps. You can use the same idea for $\mathbb R^n$ and  $\mathbb R^{n+1}$.
