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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 $\mathbb{R}^{n+1}$?

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    $\begingroup$ No nonempty subset of $\mathbb R$ is a subset of $\mathbb R^2.$ However, if $E\subset \mathbb R,$ then $E\times \{0\}$ is a subset of $\mathbb R^2.$ Apostol must be saying that if $U\subset \mathbb R$ is nonempty and open (in $\mathbb R$), then $U\times \{0\}$ is not open in $\mathbb R^2.$ $\endgroup$ – zhw. Aug 12 '16 at 3:08
  • $\begingroup$ I can't really make out your notation. $\endgroup$ – zhw. Aug 12 '16 at 3:42
  • $\begingroup$ @zhw.: I mean instead of considering $0$, can't we consider any $a\in\mathbb{R}$? More precisely, can't we interpret Apostol's statement as follows: if $U⊂\mathbb{R}$ is nonempty and open (in $\mathbb{R}$), then $U×\{a\}$ is not open in $\mathbb{R}^2$? $\endgroup$ – user 170039 Aug 12 '16 at 3:45
  • $\begingroup$ That would be equivalent, but the usual convention in considering $\mathbb R$ as a subset of $\mathbb R^2$ is $\mathbb R \sim \mathbb R \times \{0\}.$ $\endgroup$ – zhw. Aug 12 '16 at 15:17
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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 $n$.

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    $\begingroup$ thank you so much! makes sense. $\endgroup$ – Jonathan Duran Aug 12 '16 at 3:28
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the open Interval $(a,b)$ can be see in $\mathbb{R}^2$ as $$\{(x,0)\hspace{0.2cm} | \hspace{0.2cm}x \in (a,b) \}$$

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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.

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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 $\mathbb R^2$.

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Ps. You can use the same idea for $\mathbb R^n$ and $\mathbb R^{n+1}$.

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