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Find the infimum and supremum(if they exist) of the set:

$$S=\{\frac{mn}{1+m+n}: m,n\in\Bbb N\}$$

I know $$\frac{mn}{1+m+n}>0, \forall m,n\in\Bbb N$$

so $0$ is a lower bound. Now I have to find some subsequence whose limit is $0$ to prove $0$ is $\inf S$.

Let $n=\frac{1}{m}$. Then $$\lim_{m\to\infty}\frac{mn}{1+m+n}=\lim_{m\to\infty}\frac{1}{1+m+\frac{1}{m}}=0$$

So $\inf S=0$.

Is this correct? And how do I now find $\sup S$?

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    $n=\frac1{m}$, but n is a natural number. – 2017-01-15
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    oops, so I don't even know how to find $\inf S$. – 2017-01-15
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    Hint: try to work out $\inf$ and $\sup$ for $S^{-1}$ first. – 2017-01-15

2 Answers 2

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$$x=\frac{mn}{1+m+n}\implies x+1=\frac{(m+1)(n+1)}{1+m+n}\implies\frac{m+1}{m}\frac{n+1}{n}=\frac{x+1}{x}$$

Now, note that for positive integer $k$, we have $1<\dfrac{k+1}{k}\le2$.

We can thus infer that $\dfrac{x+1}{x}$ takes values between $1$ and $4$, i.e.

$$1<\frac{x+1}{x}\le4 \implies 0<\frac{1}{x}\le3\implies x\in \left[\frac{1}{3},\infty\right)$$

We then see that the infimum is $\frac{1}{3}$ and the supremum is $\infty$.

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    Why did you take 1<(x+1)\x≤4 and not 1<(x+1)\x≤2 ? – 2017-09-15
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    @Linda Because its a product of two things which are between $1$ and $2$. – 2017-09-16
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The infimum is 1. The smallest m and n for which $\frac{mn}{1+m+n}>0, \forall m,n\in\Bbb N$ is true is m=n=1, which gives you the minimum, in this case minimum =infimum. $$\frac{1*1}{1+1+1} =\frac13 >0 $$

There is no supremum. You can try to understand it by doing the following : choose the biggest n and m you can imagine, well n+1 and m+1 is going to be even bigger. This means there will always be a bigger m and n, and this is because the multiplication is stronger than the addition.

P.S.: Had to edit my answer since you can not choose m=n=0

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    You could prove it in a more formal way, but unfortunately I'm not good enough with mathjax to write it down. I hope this helped you understand why there is no supremum. – 2017-01-15
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    I am not sure that $0\in\Bbb{N}$ in OP's definition. – 2017-01-15
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    agree with @A.G. because in the second line of the original post he says that all elements of the set are strictly larger than 0 – 2017-01-15
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    Would this mean that infimum = minimum = 1 ( the smallest possible n or m in $\Bbb N$ for which $\frac{mn}{1+m+n}>0, \forall m,n\in\Bbb N$ is true)? – 2017-01-15