Showing supremum of set $S = \{1 - \frac{1}{2n}\ |\ n \in \mathbb{N} \}$

Question

I need to show the suprema of the set $S = \{1 - \frac{1}{2n}\ |\ n \in \mathbb{N} \}$ and prove my answer.

I deduced the superma to be $sup(S) = 1$.

And I prove it in two steps
$1)\ s \in S \implies \frac{1}{2} < s \leq 1 \implies 1 \geq s$, so $1$ is an upperbound.
$2)\ $By Lemma, need to show $\forall\ \epsilon > 0, \exists\ s \in S = \{1 - \frac{1}{2n}\ |\ n \in \mathbb{N} \}$ such that $s > 1 - \epsilon$.

Take $s = 1 - \frac{\epsilon}{2}$.

Is this correct?

Answer 1

You are on the right track. Personally I find it more intuitive to say that $1$ is the supremum of $S$ iff given any $\epsilon > 0$, there is an $s \in S$ so that $1-\epsilon < s \leq 1$. (That is, not only is $1$ an upper bound, but in fact it's the least upper bound.) In this case, by the Archimedean property of the real numbers, there is an $n \in \mathbb N$ so that $\epsilon > \frac 1{2n}$. So take $s = 1 - \frac{1}{2n}$. Then $1-\epsilon < 1 - \frac{1}{2n} < 1$.

Answer 2

HINT

Show that $\forall n$

$$a_n=1 - \frac{1}{2n}\le1$$

and $\forall \epsilon>0$ $\exists \bar n$ such that

$$1 - \frac{1}{2\bar n}\ge1-\epsilon$$

Answer 3

The supremum is 1.

We can show this via the following argument:

First, given that $n\in\mathbb{N}$, $\frac1{2n}>0$, so $$1-\frac1{2n}<1$$This proves the first condition for the supremum.

Second, let $k<1$. Then, let $n>\frac2{1-k}$ by the archimedian property, so$$1-k>\frac1{2n}\to1-\frac1{2n}>k$$This proves the second condition of the supremum.

Answer 4

Your proof essentially works except for the small mistake at the end, as you cannot assume that $1-\frac{\epsilon}{2}$ is in S.

What you can do is note that $\forall$ $\epsilon$ > 0 $\exists$ n $\in$ N such that $\frac1n < \epsilon$ by Archimedean property.

Then your last statement is equivalent to the statement that $\forall \frac1n > 0 $ ($n \in$ N) $\exists$ s $\in$ S such that s > $1-\frac1n$. By choosing s = $1-\frac{1}{2n}$ you can guarantee this.

Answer 5

1) $1$ is an upper bound of $S$, since

for $1-1/n \in S$:

$1-1/n <1$, $n \in \mathbb{Z^+}$.

2) Show that $\sup(S) =1$.

Assume there is a lower bound $b \lt 1$ ,

choose $\epsilon >0$ s.t. $ \epsilon < (1-b)$.

Archimedean principle:

There is a $n_0 \in \mathbb{Z^+}$ such that

$n_0 >1/\epsilon$.

For $n\ge n_0$ :

$\epsilon >1/n_0 >1/n$, and

$1-(1-b) < 1-\epsilon <$

$1-1/n_0 <1-1/n$,

$b < 1 -1/n$ .

$b$ is not an upper bound.

Hence $\sup(S)=1$..