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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 $\frac{1}{2}-\epsilon$. – RSG Jun 11 '12 at 11:10

The idea is right, the calculations I'm not that sure as I get $$\exists \epsilon >0\,\,s.t.\,\,\forall n\in\mathbb{N}\,\,,\,\frac{n}{2n+1}\leq\frac{1}{2}{-\epsilon}\Longleftrightarrow \rlap{/}{n}\leq \rlap{/}{n}+\frac{1}{2}-2n\epsilon-\epsilon\Longleftrightarrow$$$$\Longleftrightarrow2n\epsilon\leq \frac{1}{2}-\epsilon\Longleftrightarrow n\leq\frac{1}{4}\left(\frac{1}{\epsilon}-2\right)$$

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You've only dealt with part of the problem - you also need to show that $\frac{1}{2}$ is actually an upper bound for the set. It's not a problem that this part of the proof will also work for any number smaller or equal to $\frac{1}{2}$, as what you're showing is that any number smaller than $\frac{1}{2}$ cannot be the supremum, so clearly it's also true that any number smaller than $k$, for $k<\frac{1}{2}$, also can't be the supremum.

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If you really want to show the bound:

Hint: $\frac{n}{2n+1}\leq\frac{n}{2n}$

(Of course, whenever $0\leq a\leq b$, then $\frac{1}{b}\leq\frac{1}{a}$.)

If instead you're more interested in the limit,

Hint: compute $\lim_{n\rightarrow\infty}|\frac{n}{2n+1}-\frac{1}{2}|$

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First of all, check that for every $n \in \mathbb{N}$ $$ \frac{n}{2n+1} \leq \frac12. $$ This is trivial. The fact that $1/2=\sup A$ in this case follows from the fact that $$ \lim_{n \to +\infty} \frac{n}{2n+1} = \frac12. $$ This indeed implies that $\sup A$ cannot be smaller than $1/2$.

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you can use a proof by contradiction embedded in an induction proof:

Induction proof:
Check true for n=0 and n=1: 0: 0/(2x0+1) = 0; 1: 1/(2x1+1)=1/3 (yep)
Now assume true for k and check true for k+1:
So we need to show: k+1/2(k+1)+1 <= 1/2
Assume the opposite and look for contradiction:
k+1/2(k+1)+1 > 1/2
=> k+1/2k+2+1 > 1/2
=> k+1/2k+3 > 1/2
=> 2k+2>2k+3
=> k>k+1/2
Which is clearly a contradiction therefore the initial assumption is false so the oppesite is true:

k+1/2(k+1)+1 <= 1/2

so we've shown n/2n+1<=1/2 true for n=0 and n=1 and n=k+1; therefore by induction it's true for all N

The Grinch

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1. You haven't used the induction assumption... 2. This only shows $a\leq \frac{1}{2}$ for every $a \in A$. But it doesn't show that $\frac{1}{2}$ is the least upper bound (or supremum) of $A$. – Vobo Jun 14 '12 at 12:31

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