Can someone help me understand the supremum and infimum of $A = \{ \frac{n}{(n+1)} | n \in N \}$

Also $N = \{1,2,3,4...n\}$

The potential infimum and supremum I am assuming at 1 and 0 but the proof i am having trouble understanding.

I say that $|x| < M$ for some $M$ that is an upper bound so, $-M < 0 < x < 1 < M$

Now I also want to use $\alpha - \epsilon < a$, and build either a direct proof or contradiction for the supremum and $\beta + \epsilon > a$. But I can understand exactly what I'm doing wrong and what kind of conclusions to come to

I say that $1 - \epsilon < n/(n+1)$

also I say that $0 + \beta > n/(n+1)$

please help i'm getting confused

  • $\begingroup$ Your use of $n$ is a bit ambiguous. When you say $N = \{1,2,3,4...n\}$ do you mean the set is finite, with a greatest element $n$? If so, it would be better not to use $n$ as both the greatest element of $N$ AND as an arbitrary element of $N$ (as you do in writing $\{\frac{n}{n+1}|n \in N\}$) $\endgroup$ – FalafelPita Feb 13 '17 at 1:45
  • $\begingroup$ "The potential infimum and supremum I am assuming at 1 and 0...". Why supremeum smaller than infimum? $\endgroup$ – linear_combinatori_probabi Feb 13 '17 at 1:46

For the supremum

$\frac{n}{n+1} < 1$ for all $n$, so the supremum can't be greater than 1. Let $a$ be the supremum, and suppose to the contrary that $a < 1$. Then for any $m$ such that $\frac{a}{1-a} < m$ we have that $a < \frac{m}{1+m}$, so $a$ is not an upper bound of $A$. Contradiction. Hence, $a \geq 1$. So $a =1$.

For the infimum

According to your definition of $A$, the infimum would not be $0$, because you said $n$ must be a positive integer. So $A$ has a least element, which is $\frac{1}{1+1}$.


For positive integers $n$, define $x_n = \frac{n}{n+1}$ and note that $A = \{x_1,x_2,...\}$. Now,

$(n+2)n = n^2+2n < n^2 + 2n + 1 = (n+1)(n+1)$

$\Longrightarrow x_n = \frac{n}{n+1} < \frac{n+1}{n+2} = x_{n+1}$.

As an increasing sequence, if $(x_n)_{n=1}^\infty$ converges at all, its limit must be $\sup_n x_n = \sup A$. This sequence does in fact converge, and $\lim_{n \to \infty} x_n = 1$, which tells you that $\sup A = 1$. To see why the sequence converges, just rewrite:

$x_n = \frac{\frac{1}{n}}{\frac{1}{n}}\cdot \frac{n}{n+1} = \frac{1}{1 + \frac{1}{n}} \to \frac{1}{1+0} = 1$.

Next, as for the set $N$ you describe above, it is finite, so its supremum is just its largest member.


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