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$$ A=\left\{ x \in \mathbb{R}^{p} \mid \forall i:\ x_{i} \in (-1,1) \right\}$$

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Pick $x\in A$ at random, and choose $\delta = \min B$ where: $$ B = \{ 1-x_{i}\mid i \in \{1,2,\dotsc,p\} \} \cup \{ 1+x_{i}\mid i \in \{1,2,\dotsc,p\} \} $$ Now pick $y\in \{ z\in \mathbb{R}^p \mid \|x-z\| < \delta \}$. I have to show that the components $y_i$ are strikly greater than $-1$ and striktly smaller than $1$.

This is where I'm stuck. The only thing I know is this: $$ \sqrt{\sum_{i=1}^p(x_i-p_i)^2} < \delta $$

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$$\|x-y\|<\delta \implies \sqrt{\sum_{i=1}^{n}(x_i-y_i)^2}<\delta\implies|x_i-y_i|<\delta\implies|y_i-x_i|<\delta \tag{1}$$ Now, $$(\delta\le1-x_i) \land (\delta\le1+x_i)$$ $$\implies (x_i\le1-\delta) \land (-x_i\le1-\delta)$$ $$\implies |x_i|\le 1-\delta \tag{2}$$ Using $(1)$ and $(2)$ and triangle inequality: $$|y_i|\le|y_i-x_i|+|x_i|<\delta+(1-\delta)=1\ \square$$

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We denote $A_{i}=(-1,1)$ for $1\leq i\leq p$ $$ A=\Pi_{i=1}^{p}A_{i} $$

then if $(x_{i})_{i=1}^{p}\in A$ then $x_{i}\in A_{i}$ but $A_{i}$ is open so there is an open ball $B(x_{i},\epsilon_{i})\subseteq A_{i}$ so by letting $\epsilon:=\min\{\epsilon_{i}\mid1\leq i\leq p\}$ we get $$ B(x_{i},\epsilon)\subseteq B(x_{i},\epsilon_{i})\subseteq A_{i} $$

so that $$ B(x,\epsilon)\subseteq A $$

so $B$ is an open neighborhood of $x$ contained in $A$, thus $A$ is open.

Note that I have only used the fact that each $A_i$ is open to prove their product is open.

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  • $\begingroup$ Isn't a finite product of open sets - in the producttopology - open by definition? I somehow dislike the term "prove" in your last line. $\endgroup$ – drhab Apr 25 '15 at 11:52
  • $\begingroup$ @drhab - It is prove! If $X=\Pi_{i=1}^{n}\mathbb{R}$ one can consider two topologies on $X$ - The product topology and the topology induced by the metric on $\mathbb{R}^{p}$. I proved that $A$ was open without relying on the fact that the two topologies on $X$ are the same since I assumed the OP does not know that $\endgroup$ – Belgi Apr 25 '15 at 12:06
  • $\begingroup$ That is a convincing answer to my comment. I will delete it. $\endgroup$ – drhab Apr 25 '15 at 12:25
  • $\begingroup$ @drhab - I think you should keep it, it can be interesting for future readers as it is a confusing matter $\endgroup$ – Belgi Apr 25 '15 at 12:29
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Not really an answer to your question, but a suggestion to follow a different and in my view more elegant route (too much for a comment).

Denote $A_i=\{x\in\mathbb R^p\mid x_i\in (-1,1)\}$ for $i=1,\dots,p$.

If I understand well then it is your goal to prove that $A=\bigcap_{i=1}^pA_i$ is open in $\mathbb R^p$. For this it is enough to prove that the $A_i$ are open. In that case $A$ is open as finite intersection of open sets.

Actually the $A_i$ are open by definition when every copy of $\mathbb R$ is equipped with its usual topology and $\mathbb R^p$ is equipped with the producttopology. This because $(-1,1)\subset\mathbb R$ is an open set, and sets of the form $\{x\in\mathbb R^p\mid x_i\in U\}$ with $U$ open in $\mathbb R$ form a subbasis of the producttopology.

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