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In a normed vector space $E$ the only open and closed set is $E$ and the empty set as a corollary to connected properties. Also an closed ball cannot be open, I would like to prove this result with only the fact that :

A set $O$ of a normed vector space $(E,\Vert\Vert)$ is open if and only if $$\forall x\in O\quad \exists\varepsilon>0\quad B(x,\varepsilon) \subset O.$$

So if the closed ball $B(a,r]=O$ is open whenever I choose an element of $O$ there exist $\varepsilon>0$ such that $B(x,\varepsilon)\subset O$. With a picture is cleary false by choosing an element $x$ such that $\Vert x-a \Vert=r$. I have to choose $x=a+r\nu$ where $\nu$ is an unitary vector. Then I have to prove that the open ball centered at $a+r\nu$ with a strictly positive radius $\varepsilon$ cannot be included in $O$.

Am I right?

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  • $\begingroup$ @mookid please do not change my notation. $\endgroup$ – user169373 Nov 3 '14 at 22:33
  • $\begingroup$ okay. I thought it was a typo. $\endgroup$ – mookid Nov 3 '14 at 22:34
  • $\begingroup$ @mookid no problem, does my vector is correct? $\endgroup$ – user169373 Nov 3 '14 at 22:55
  • $\begingroup$ @MarcoGato I made this an answer actually. $\endgroup$ – mookid Nov 3 '14 at 23:04
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This is a good start. A final argument argument would to say that for any vector $y$ with $|x−y|=r$, for any $ϵ>0$, $$y_ϵ=y+\frac{ϵ}{2r}(y−x)$$ [$y_\epsilon$ is just a little bit further from $x$ in the $y$ direction] is such as $|y−y_ϵ|<ϵ$ but $|y_ϵ−x|>r$.

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You want to prove that $B[a,r]$ is not open. You said ot yourself: take $x \in B[a,r]$ such that $\|x - a\| = r.$ Take any $\epsilon > 0$. Then the ball $B(x,\epsilon)$. Is not contained in $B[a,r]$. Take a convenient vector and use the triangle inequality.

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  • $\begingroup$ Thanks but you don't answering my question, in other words does my vector $x=a+r\nu$ was correct? $B(a+r\nu,\epsilon)$ $\endgroup$ – user169373 Nov 3 '14 at 22:36
  • $\begingroup$ Yes, you are correct. You have to take a vector in that last ball, and use the triangle inequality to show that it is not in $B[a,r]$. $\endgroup$ – Ivo Terek Nov 3 '14 at 23:03

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