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.

  • 1
    $\begingroup$ The supremum is the least upper bound. In this case it's very easy to guess it, if you think about the structure of your set. Once you have identified the supremum (let's call it $S$), you have to prove that it's the least upper bound possible, that is that for every $\varepsilon > 0$, there exists an $n \in \mathbb N$ such that $x_n > S - \varepsilon$. $\endgroup$
    – rubik
    Nov 5, 2017 at 15:46

1 Answer 1


The least upper bound (i.e., the supremum) is $1$.


$1$ 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.


When proving a number $a$ is a supremum for a set $S$, you have to do two things:

  1. Show that $a$ is an upper bound for $S$.
  2. Prove that for any other upper bound $b$, $a \le b$.
  • $\begingroup$ You can start by placing the math inside dollar signs ($...$ for inline math, $$..$$ for display style math). In this case it should be sufficient. $\endgroup$
    – rubik
    Nov 5, 2017 at 15:49
  • $\begingroup$ Thanks for the help @rubik $\endgroup$
    – ATOMP
    Nov 5, 2017 at 16:10
  • $\begingroup$ Good job, it's much more readable now. Also, +1. $\endgroup$
    – rubik
    Nov 5, 2017 at 20:54

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