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Please just don't present a proof, see my reasoning below

I need to find the sup and inf of this set:

$$A = \{\frac{n+1}{n}, n\in \mathbb{N}\} = \{2, \frac{3}{2}, \frac{4}{3}, \frac{5}{4}, \cdots\}$$

Well, we can see that:

$$\frac{n+1}{n} = 1+\frac{1}{n} > 1$$

Therefore, $1$ is a lower bound for $A$, however I still need to show that it's the greatest lower bound of $A$. Suppose that $1+\frac{1}{n}>c>1$, then $\frac{1}{n}>c-1\implies \frac{1}{c-1}>n\implies$ c has to be $>1$, which is not a problem :c.

Now, for the sup, we have:

$$\frac{n+1}{n} = 1+\frac{1}{n}\le 2$$

because if $n>1$ then $\frac{1}{n}<1$ then $1+\frac{1}{n}<1+1=2$. So, $2$ is the upper bound of $A$, but I still have to show that $2$ is the lowest upper bound of $A$. I've read that if $a\in A$ and $a$ is an upper bound, then $a$ is the sup (how do I prove it?). But suppose that I didn't knew this theorem, then I would have to prove that there is no $c$ such that

$$1+\frac{1}{n}<c<2$$

and such that $c\ge a, \forall a\in A$. Oh, wait, I might have been able to prove the question above: suppose $c\ge A, \forall a\in A$, with $c\in A$. Then if there exists another $b$ such that $c>b\ge a$, i can see that this $b$ is greater than every member of $A$, but not $c$, therefore there isn't such $c$ that is both greater than every member of $A$ and in the middle of $c$ and $a$.

Moishe Kohan
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Guerlando OCs
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1 Answers1

4

Answer on "how do I prove it?"

If $a\in A$ is an upper bound of $A$ then any $c$ with $c<a$ is not an upper bound of $A$ since $a$ is an element of $A$ that does not satisfy $a\leq c$.

We conclude that $a$ must be the least upper bound of $A$.

  • $1\leq1+\frac{1}{n}$ for each $n$ so $1$ is a lower bound of $A$.

  • If $c>1$ we can find an element $1+\frac{1}{n}\in A$ with $1+\frac{1}{n}<c$ so if $c>1$ then it is not a lower bound of $A$.

This together justifies the conclusion that $1=\inf A$.

  • $1+\frac{1}{n}\leq2$ for each $n$ so $2$ is an upper bound of $A$.

  • If $c<2$ then we have element $2\in A$ with with $c<2$ so if $c<2$ then it is not an upper bound of $A$.

This together justifies the conclusion that $2=\sup A$.

drhab
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