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I'm attempting to prove rigurously that the hypercube is an open set in $\mathbb{R}^p$.

The problem states.

Let $A= \lbrace (x_1,x_2,...,x_p) \in \mathbb{R}^p : 0<x_1<1 , 0<x_2<1 , \dots , 0<x_p<1 \rbrace $. Prove $A$ is open, using the usual metric.

So far, I was doing this.

I have to prove that $\forall x \in A $ , $\exists r>0$ such that $B_r(x) \subseteq A$.

That is

if $x\in \mathbb{R}^p$ then exists $r>0$ such that $B_r (x) \subseteq A$ and if $z\in B_r (x)$ then $z\in A$.

So I proposed $r=1-\max\lbrace x_1,x_2,x_3,...,x_p\rbrace $. That means that $1-\max\lbrace x_1,x_2,x_3,...,x_p\rbrace <x_i $ $\forall i=1,2,...,p$.

Also if $z=(z_1,...,z_p)\in B_r (x)$ then $|z_i-x_i|<d(z,x)<r=1-\max\lbrace x_1,x_2,x_3,...,x_p\rbrace<x_i $.

So:

$|z_i-x_i|<x_i$ then $-x_i<z_i-x_i<x_i$ for all $i=1,...,p$. Then:

$0<z_i<2x_i<2 $ , $\forall i=1,...,p$. But i wanted $0<z_i<1$ for all $i=1,...,p$ so $z$ would be in $A$. What I've done wrong so far?

Thanks in advance

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  • $\begingroup$ Look at your $r$: it might be negative. Instead, consider $r := \frac{\min_i x_i}{2}$ $\endgroup$ – Hermès Oct 5 '16 at 7:12
  • $\begingroup$ AHHH Sorry, i screwed it up, i'll correct the statement $\endgroup$ – User117E29 Oct 5 '16 at 7:13
  • $\begingroup$ Now it's the correct exercise :) $\endgroup$ – User117E29 Oct 5 '16 at 7:14
  • $\begingroup$ What if $x=(\epsilon,\ldots,\epsilon)$ for $\epsilon>0$ sufficiently small? $\endgroup$ – Paolo Leonetti Oct 5 '16 at 7:19
  • $\begingroup$ Ohh, i see, In $\mathbb{R}^2$ the radius would be "big" $\endgroup$ – User117E29 Oct 5 '16 at 7:24
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Fix $x \in \,]0,1[^p$ and consider the ball $B_r(x)$ with radius $$ r:=\min\{\min\{x_i,1-x_i\}: i=1,\ldots,p\}. $$ It follows that, if $z \in B_r(x)$ then, for each $i=1,\ldots,p$, it holds $$ z_i \in \,\,]x_i-r,x_i+r[ \,\,\subseteq \,\,]0,1[\,. $$

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  • $\begingroup$ So it was tricky. Thanks a lot! I just can't figure it out why it's the max $\endgroup$ – User117E29 Oct 5 '16 at 7:40
  • $\begingroup$ wouldn't be min{min .... or what i'm not seeing $\endgroup$ – User117E29 Oct 5 '16 at 7:43
  • $\begingroup$ Obviously, I had to finish my sleep ;) $\endgroup$ – Paolo Leonetti Oct 5 '16 at 7:44
  • $\begingroup$ Lol, thanks a lot! :) I'm in Mexico, it's 2:44 am, i should sleep... $\endgroup$ – User117E29 Oct 5 '16 at 7:45
  • $\begingroup$ The right time to think about open balls :D good night then! $\endgroup$ – Paolo Leonetti Oct 5 '16 at 7:46

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