I've worked through a proof for this set and I don't know how to check my correctness, would also be thrilled if anyone has insights on more efficient ways to do these proofs that might apply more generally to proving the sup and inf of sets!

Proof: Let $S = \lbrace\frac{2}{n}\mid n \in\lbrace {1, 2, 3, 4,...}\rbrace\rbrace$

Notice: $ 0 \leq \frac{2}{n} \leq 2$

Therefore, S is bounded and a sup and inf exist.

$Sup(S) = 2$ trivially since $\frac{2}{1} \geq \frac{2}{n}\ \forall\ n\ \in S$

$0$ is a lower bound for S.

$0 \leq inf(S)$ and sidenote ($\lim_{x\to\infty} \frac{2}{n} = 0$)

Now, suppose there is an $inf(S) > 0$

Choose $\frac{1}{inf(S)}$ where $inf(S) \neq 0$

By the Archemedian Property, there exists a $n \in N$ such that $n > \frac{1}{inf(S)}$

Therefore, $inf(S) > \frac{1}{n} > 0$ by $x>y>0$ and $\frac{1}{y}>\frac{1}{x}>0$

But now $\frac{1}{n} \in S$ and $inf(S) > \frac{1}{n}$. Contradicting n being a GLB of S.

Therefore, $inf(S) = 0$

Comments: Firstly, sorry for the horrid formatting, I've tried to clean it up a little but I can understand how it still looks cluttered. I'm just not really sure if I am safe to assume inf(S) = 0 with my conclusion. I think this works with $\frac{1}{n}$ as the set operation but less sure with $\frac{2}{n}$

I hope my working is readable and thank you for the responses in advance. Here and ready to read and reply.


Your work is fine.

Let me suggest:

$0\lt a_n:= 2/n$ , $n \in \mathbb{Z^+}.$

Hence $0$ is a lower bound of $S .$

Need to find $\inf(S)$, the greatest lower bound.

Assume $B>0$ is a lower bound.

Archimedes :

There is a $n_0$ such that $n_0 \gt 2/B$.

For $n \ge n_0 :$

$2/n \le 2/n_0 \lt B .$

Hence $B$ is not a lower bound :

$\rightarrow \inf(S)=0.$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.