I was thinking that i could prove that all elements in $\mathbb R^n-(0,0,...,0)$ are interior points. That meaning if given any $p\in \mathbb R^n$ there exists $\epsilon>0$ such that $B_{\epsilon}(p)\subset \mathbb R^n-(0,0,...,0)$. I thought of choosing $\epsilon=d(p,0)/2$, but I'm not quite sure of the answer. Any suggestions or solutions?
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If your aim is to show that $R^n \setminus {(0,0,...,0)}$ is open, you can think in this way : as ${(0,0,...,0) }$ is closed in $R^n$, then its complement is open, so that ${(0,0,...,0) } ^c$ is open . – Nizar Oct 01 '15 at 06:47
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You just gave me an idea. Another way to prove the proposition is to show that ${(0,0,...,0)}$ is closed proving that the origin is the only accumulation point of $\mathbb R^n{(0,0,...,0)}$. If I take $p\in \mathbb R^n{(0,0,...,0)}$ and make $\epsilon = d(p,0)+1$ then $B_{\epsilon}(p)$ will contain the origin, thus making the ball a non open set. Im not sure if it helps – Raúl Oct 01 '15 at 07:48
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What part are you not sure about? You should show it's impossible for 0,0... to be in that ball, if you choose epsilon small enough. It looks like you're on the right track so far.
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I guess you want something more rigorous to justify the intuitively obvious choice of radius. This can be done: Let $d$ be the usual metric; let $x \in \mathbb{R}^{n} - \{ 0 \}$; then $$ d(y,0) \geq d(x,0) - d(x,y) > 0 $$ if $d(x,y) < \frac{d(x,0)}{2}$, by the triangle inequality and the symmetry of metrics; therefore, the open ball $B^{x}(\frac{d(x,0)}{2})$ of center $x$ and radius $\frac{d(x,0)}{2}$ is included in $\mathbb{R}^{n} - \{ 0 \}$, and we are done.
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