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 Post subject: Open subset
PostPosted: Thu Oct 05, 2017 10:02 am 
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 423
Let \(\displaystyle{G}\) be a non-empty and open subset of \(\displaystyle{\left(\mathbb{R},|\cdot|\right)}\)

such that \(\displaystyle{x\pm y\in G\,,\forall\,x\,,y\in G}\). Prove that \(\displaystyle{G=\mathbb{R}}\).


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 Post subject: Re: Open subset
PostPosted: Thu Oct 05, 2017 9:49 pm 

Joined: Sat Aug 12, 2017 12:12 pm
Posts: 4
Since $G$ is non empty, let $a \in \displaystyle{\mathbb{R}}$ be a real that belongs to $G$. But then, $a-a$ must also belong to $G$, that is, $0 \in G. \\$
Now since $G$ is open, there must exist an open interval around $0$, say $(-\delta,\delta) \subseteq G$ with $\delta >0$. Now pick any positive real $\theta. \\$
Choose a big enough $n \in \displaystyle{\mathbb{N}}$ such that $\theta / n < \delta$. This means that $\theta / n \in G$. By adding this number to $0 \in G, n$ times, you $\\$
get (every time) an element of $G$. Hence, $\theta \in G. \\$
If a negative real $t$ is, futhermore, selected, then $-t$ is positive and according to the above, belongs to $G$. But then,
$0 - (-t) = t$ must also belong to $G$. Hence, all positive and negative reals, and $0$, belong to $G$. Hence, $G = \mathbb{R}$



Efthymios Tsakaleris


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 Post subject: Re: Open subset
PostPosted: Sat Oct 07, 2017 12:53 pm 
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Thank you.


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 Post subject: Re: Open subset
PostPosted: Thu Oct 12, 2017 8:07 am 

Joined: Sat Aug 12, 2017 12:12 pm
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Welcome


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