Is this sufficient? Also, any good books/other suggestions regarding the subject will be very helpful.

Find min, max, inf, sup (if they exist):


Showing B has an upper bound: Let $M=1$, we need to find $m,n$ fulfilling:$$\frac{m}{m+n}>1$$ As $n\in\mathbb{N}$ and is only in the denominator, the smaller it's value, the greater the value of n, the smaller $b$ will be. Therefore, let us choose $n=1$ (smallest possible value).$$\frac{m}{m+1}>1\,\,\,\,\,\leftrightarrow\,\,\,\,\,\,m>m+1$$

We got a contradiction, thus $M$ is an upper bound of $B$.

Showing $M=\sup B$: Let $\epsilon>0$, we need to find $b\in B$ fulfilling:$$\frac{m}{m+n}>1-\epsilon$$ Again, we'll choose $n=1$ to get the biggest $b$ possible: $$\begin{align} \frac{m}{m+1}&>1-\epsilon\\ m&>m+1-m\epsilon -\epsilon\\m&>\frac{1-\epsilon}{\epsilon} \end{align}$$ Therefore for every $\epsilon$ we can choose $n=1,m>\frac{1-\epsilon}{\epsilon}$, which means $\sup B=1$.

Edit: Since $m,n \in\mathbb{N}$, $B>0$.

Showing $0=\inf B$: Let $\epsilon>0$, we need to find $b\in B$ fulfilling: $$\frac{m}{m+n}<0+\epsilon$$ Choosing $m=1$ to make $b$ as small as possible: $$1<\epsilon+n\epsilon\\n>\frac{1-\epsilon}{\epsilon}$$

We have shown that such $b$ exists for every $\epsilon$. Therefore, $\sup B = 0$

  • 1
    $\begingroup$ $n$ and $m$ are natural numbers. You seem to have forgotten that. The supremum is one and the infimum is zero. $\endgroup$
    – Pedro
    Commented Nov 12, 2012 at 13:50

3 Answers 3


Your proof that $1$ is an upper bound is unnecessarily complicated: as $m,n>0$, we have $m<m+n$, and then $m/(m+n)<1$.

Also, as was mentioned, $0$ is a lower bound (since everything is positive). And it is the infimum, as $1/(1+n)\to 0$.

  • $\begingroup$ Just to add to Martin's answer. An upper bound is the supremum if and only if there exists a sequence inside the set that converges to it. The OP has showed that $1$ is an upper bound, so we only need to find one sequence in the set convergent to it. This can be done by simply setting $n=1$ and taking $m\geq 1$, i.e. $x_{m}:=\frac{m}{m+1}$, and it converges to $1$. Similarly for inf. $\endgroup$ Commented Sep 13, 2023 at 12:15

If you take $\,\Bbb N=\{1,2,3,...\}\,$, then I think you'll agree with

$$\forall\,\,m,n,\in\Bbb N\,\,\,,\,\,\frac{m}{m+n}>0\Longrightarrow 0\,\,\text{is a lower bound for}\,\,M\,...$$

I think it'd be a good idea to try to prove that zero is actually the infimum of $\,M\,$


Consider first case where the value m is much larger than the value of n. $n << m$

Then consider the case where the value n is much larger than the value m. $m <<n$

Write out a few iterations and you'll see where each one is headed. That will give you the supremum and infimum.

I should point out like the other posts that $m,n \in \mathbb{N}$ which means $m,n > 0$.


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