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Let $A=\left\{\dfrac{n}{2n+1}:n \in \mathbb{N}\right\}$. I want to prove that $supA=\dfrac{1}{2}$ so I need to show that $$\forall\epsilon\gt0 \exists a\in A:a\gt\dfrac{1}{2}-\epsilon$$

So suppose by contradiction that $$\exists \epsilon\gt0 \forall a\in A:a\le\dfrac{1}{2}-\epsilon$$

which essentially means that $$n \le ( \dfrac{1}{\epsilon}-1)\cdot \dfrac{1}{2}$$ and this is not possible as the natural numbers are not bounded from above.

Am I right? This seems to easy - I could have plugged any other number (smaller then $\frac{1}{2}$) and prove the same.

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    Just rewrite the whole logic again without the contradiction method, i.e. fixing an $\epsilon$ find an element greater than $\frac{1}{2}-\epsilon$. You can show by the same logic that for any $n\geq\left[\left(\frac{1}{\epsilon}-1\right)\frac{1}{2}\right]+1$ all the members of the set will be greater than $\frac{1}{2}-\epsilon$.2012-06-11

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