# Prove that the hypercube is an open set

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?

• Look at your $r$: it might be negative. Instead, consider $r := \frac{\min_i x_i}{2}$ – Hermès Oct 5 '16 at 7:12
• AHHH Sorry, i screwed it up, i'll correct the statement – User117E29 Oct 5 '16 at 7:13
• Now it's the correct exercise :) – User117E29 Oct 5 '16 at 7:14
• What if $x=(\epsilon,\ldots,\epsilon)$ for $\epsilon>0$ sufficiently small? – Paolo Leonetti Oct 5 '16 at 7:19
• Ohh, i see, In $\mathbb{R}^2$ the radius would be "big" – User117E29 Oct 5 '16 at 7:24

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 $$z_i \in \,\,]x_i-r,x_i+r[ \,\,\subseteq \,\,]0,1[\,.$$