TITLE: Supremum derivation
QUESTION [0 upvotes]: This is the question I am trying to complete:
Find the supremum of the following set
$$M:= \left\{x \in \mathbb R | x= 1-\dfrac{1}{n} , n \in \mathbb N\right\}$$
I know that the supremum is an upper bound and the supremum of that function but I don't know how to find a suitable derivation/proof for it.

REPLY [2 votes]: The least upper bound (i.e., the supremum) is $1$.
Proof
$1$ is an upper bound for the set, since $1 - \frac{1}{n} < 1$  for all $n > 0$.
Assume by way of contradiction that $$b = 1 - e, \space e \in \Bbb R$$ is another upper bound for $M$ that is less than $1$. 
But by the Archimedian property of rational numbers, there exists a natural number $m$ where $\frac{1}{m} < e$ (Archimedean Property and Real Numbers). 
But $1 - \frac{1}{m} \in M$, and $1 - \frac{1}{m} > b$. So $b$ is not an upper bound for $M$, which is a contradiction.
Therefore, $1$ is the least upper bound (i.e., the supremum) of the set.
Note
When proving a number $a$ is a supremum for a set $S$, you have to do two things:

Show that $a$ is an upper bound for $S$.
Prove that for any other upper bound $b$, $a \le b$.
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