Is this a valid way to prove a beginer supremum question

Question

Hello suppose I have a set $$S=\left\{1-\frac{1}{n} : n \in \mathbb{N}\right\}$$

and I want to show that $\sup(S)=1$. Is the following valid?

I see that $S$ is bounded from above , in particular by $1$ because $\frac{1}{n}$ never is negative.

Now suppose $\sup S=1$

now must show it is the least upper bound,

The Archimedian property tells us that for any $\epsilon \in \mathbb{R}$ there exists a $n_{\epsilon} \in \mathbb{N}$ such that $\epsilon \lt n_{\epsilon}.$

So $\frac{1}{\epsilon} \gt \frac{1}{n_{\epsilon}}$ and so $1-\frac{1}{\epsilon} \lt 1-\frac{1}{n_{\epsilon}}$ and $1-\frac{1}{n_{\epsilon}}$ is certainly in S so I can conclude indeed 1 is the least upper bound.

This is my first times working with these types really, so I am not sure if it is valid. Hopefully someone can conform is so, and if not please help me understand any mistakes.

Thanks

Answer 1

In your answer, you must show for any $\epsilon > 0$, you can find an $n$ such that $1-\epsilon < 1-\dfrac{1}{n}$, and this means $n > \dfrac{1}{\epsilon}$. Thus choose $n > \dfrac{1}{\epsilon}$ is sufficient.

Answer 2

You correctly shows that $1$ is an upper bound for $S$. To complete the proof that $1$ is the least upper bound for $S$, you need to show tha tthere is no upper bound strictly smaller than $S$. You can do so by contradiction: Assume $a$ is an upper bond for $S$ and $a<1$.

Recall that we want to exhibit and element $1-\frac1n$ of $S$ that is $>a$. For this it suffices to have $n>\frac1{1-a}$.

As $a<1$, we have $1-a>0$ and hence $\epsilon:=\frac1{1-a}>0$. By the Archimedean property, there exists $n\in\mathbb N$ with $n>\epsilon$. Then $\frac1n<1-a$ and $1-\frac1n>a$ as desired. We conclude that no upper bound is $<1$ and so $1$ is the least upper bound.