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My answer is not quite robust enough, but this is what I'm thinking... Think I can state it clearly for any number $c \geq 1$ but I know that doesn't cover any number $0\lt c \le 1$

let $c$ be a positive number. Prove that the set $S$ = {$c, 2c, 3c, ..., nc, ...$} is not bounded above

Let's assume that $c$ is equal to $1$. In the set $S$, we can see that the result would be $S$ = {$1, 2, 3,..., n,...$} which has no upper bound.

Since any positive number $c > 1$ would behave the same way, we can see there would be no upper bound in this set for any number $c$ $\geq 1$

Naturally this would also hold true for any number greater than $0$

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Maybe another way is to take a look at

$f \colon \mathbb{N} \rightarrow S$ with $f(n) = cn$. Then obviously $\mathrm{Im} f = S$, and you only need to show that $f$ is unbounded.

So assuming there is an $M \in \mathbb{R}$ s.t. $$f(n) \leq M \quad \forall \, n \in \mathbb{N}$$.

Then $cn \leq M$ for all $n \in \mathbb{N}$ which is equivalent to $c \leq \frac{M}{n}$ for all natural numbers.

Choosing now $n \geq \frac{2M}{c}$ we observe that $c \leq \frac{c}{2}$ must hold, which is a contradiction.

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Let $M$ be any positive real number, and note that $M \leq c\lceil{M/c}\rceil$, where $\lceil{-}\rceil$ is the ceiling function, so $c\lceil{M/c}\rceil \in S$. That is to say, for any positive real number $M$ we can find a number $c\lceil{M/c}\rceil$ in $S$ greater than $M$, so $S$ is not bounded above.

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Let $k$ denote the least upper bound so some $n\in\Bbb N$ satisfies $nc>k-c$, whence $(n+1)c=nc+c\ge k$, a contradiction.

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  • $\begingroup$ I'm having some trouble following this one (my apologies). Would something like this work: $\frac{nc}{(n+1)c} \gt 1$? $\endgroup$
    – B. Hoffman
    Oct 24, 2019 at 2:54
  • $\begingroup$ @B.Hoffman I don't see how you derived it. The point is if $k$ is minimal any real $<k$, including $k-c$, is exceeded by some element of $S$. $\endgroup$
    – J.G.
    Oct 24, 2019 at 5:29

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