Analysis proof involving bounded intervals

Question

This proof is somewhat similar to the question asked here: Help with real analysis proof involving supremum. But it's the other way around.

Question: Let $S$ be a non-empty subset of the real numbers, bounded above. Show that if $u = sup S$, then for every natural number $n$, the number $u − 1/n$ is not an upper bound of $S$, but the number $u + 1/n$ is an upper bound of S.

Here is my approach. First, I write the conditions for my proof to satisfy. They are, for every natural number $n$ the following two must satisfy,

$$ (I)\hspace{1cm} s\leq u+1/n,\forall s\in S \\ (II)\hspace{1cm}\exists s \in S \ni s\geq u-1/n $$

Now we prove this by induction. First, we have to establish the base case for $n=1$.

We can easily show that, when $n=1$

$u+1/n = u+1$

Since $u+1>u$, it is clearly an upper bound for $S$. (I) holds.

and $u-1/n = u-1$.

since $u-1 < u$, it is clearly not an upper bound for $S$. Which means (II) holds.

$\\\\\\$

Now let's assume that the conditions (I) and (II) are true for some integer $m$.

i.e:

$ s\leq u+1/m,\forall s\in S$ and

$\exists s \in S \ni s\geq u-1/m$

but the case $m+1$ implies, $u+1/m > u+1/(m+1)$

It can be proven using the following process.

$u+1/m > u+1/(m+1) \Leftrightarrow 1/m > 1/(m+1) \Leftrightarrow m+1 > m \Leftrightarrow 1>0 $

Now the problem is this does not guarantee $ s\leq u+1/(m+1),\forall s\in S$.

Note that $u+1/(m+1)$ fall between $u$ and $u+1/m$.

The same is true for $u-1/(m+1)$ situation.

Any insights into this would be appreciated.

Answer 1

Supremum means the least upper bound. You are given that u is the least upper bound of the set S. $u+1/n>u$ for all natural numbers $n$ since $1/n>0$. So, it is greater than the least upper bound, meaning it is an upper bound. Similarly, $u-1/n<u$ since $1/n<0$ for all natural numbers $n$. Since it is smaller than the least upper bound, it is not an upper bound.

Answer 2

1. Let $T$ be the set of all upper bounds for $S.$ By definition, $u=\sup S$ is the least member of $T.$ If $n\in \Bbb N$ then $u-1/n<u.$ So if $u-1/n$ were an upper bound for $S$ then $u-1/n$ would be a member of $T$ that's less than the least member ($u$) of $T$, which is absurd.

2. By definition of $T,$ we have, for any $v,v',$ that $$v'>v\in T \implies \forall s\in S\,(v'>v\ge s)$$ $$ \implies \forall s\in S\,(v'\ge s)$$ $$\implies v'\in T.$$ In particular $n\in \Bbb N\implies u+1/n>u\in T\implies u+1/n\in T.$

Answer 3

Note: $u - \frac 1n < u = \sup S$. And note that $u + \frac 1n > u = \sup S$.

That is IT! You are done.

The definintion of $u = \sup S$ is:

1. $u\ge x$ for all $x \in S$.

So $u+ \frac 1n > u \ge x$ for all $x \in S$. So $u+\frac 1n$ is an upper bound. That's all there is to it.

2. If $w < u$ then $w$ is not an upper bound of $S$.

So $u-\frac 1n < u$. So $u-\frac 1n$ is not an upper bound of $S$. End of story.