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I'm trying to prove that $\exists A > 0$ such that $a_n+A > 0, ∀ n ∈ \mathbb{N}+$ for a bounded sequence $(a_n)_{n∈\mathbb{N}+}$. My proof so far is as follows:

If $(a_n)_{n∈\mathbb{N}+}$ is bounded $\Rightarrow$ $\exists U∈\mathbb{R}$ such that $a_n\ge U, \forall n∈\mathbb{N}+$. We set the minimum element in the set to equal $U$: min{$a_n$}=$U$.

We want to show $U+A\gt0 \Rightarrow a_n+A \gt 0, \forall n∈\mathbb{N}+$.

Because $U+A\gt0$ holds, then $A>-U$. Because $-U∈\mathbb{R}$, $\exists A$ which satisfies $A>-U$

Therefore, $\exists A > 0$ such that $a_n+A > 0, ∀ n ∈ \mathbb{N}+$.

Is this proof correct? Any help would be appreciated.

aL_eX
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    If $U+A>0$ then $a_n+A>0$ follows directly from $a_n\geq U$ (so e.g. $A:=1-U$ will do the job). Btw, you should say that $U=\inf{a_n\mid n\in\mathbb N_+}$. Not minimum. It is possible that the set doesn't even has a minimum. – drhab Oct 30 '16 at 13:57
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    You invoked the Archimedean property, unknowingly, when you wrote, −U∈R, ∃A which satisfies A>−U . (If you assumed A to be a natural number) – IamThat Oct 30 '16 at 13:59
  • I think it's simpler to just set $A=|I|+1$ where $I$ denotes the infimum of the sequence, which exists and is finite. – learner Oct 30 '16 at 14:03

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Yes, it is correct, though you could simply say:

  • Let's $U<a_i,$ $\forall i\in \mathcal{N}$.
  • Take $A = 1-U$. Then, $$a_n + A >U+1-U=1>0$$
sdd
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