Suppose $S$ is a non-empty subset of $\mathbb{R}$ bounded below.

Prove that: If $L$ is a lower bound of $S$, then $L= \inf(S)$ iff $\exists$ a sequence $\{S_n\}$ of elements of $S$ that converges to $L$


($\Longrightarrow$) Assuming that $\inf(S) = L$. I know that I have to construct a sequence that converges to $L$. This being the case we know by the assumption that: $$\forall \epsilon > 0\ \exists\,s_n\in S\text{ s.t. }L < s _n< L+\epsilon$$

And what I want to show is that: $$\forall \ \epsilon > 0 \ \exists \ N \in \mathbb{N} \ s.t. \ if \ n\geq N \ then \ |s_n - L| < \epsilon$$

So to construct my sequence:

Let $S_n$ = $\{s_i \in S| s_i \notin (L,L+\epsilon)$ and let the terms be defined as follows:

We pick a point $s_1 \in (L,L+\epsilon)$ that exists based on our assumption. Then we decrease the size of $\epsilon$. If $s_i \in (L,L+\epsilon)$ we decrease the size of $\epsilon$ until $s_i \notin (L,L+\epsilon)$. Once this is the case we label that value as a term of our sequence $s_n$. We then repeat this process again, we pick a new $s_i \in (L, L+\epsilon)$ which is now a smaller interval. We decrease the size of $\epsilon$ again and if our newly selected $s_i \notin (L,L+\epsilon)$ we add it to our sequence.

We are guaranteed the continued existence of these $s_i$ as we decrease $\epsilon$ because of the Archemdian property. In this way we have constructed a sequence and as $\epsilon \rightarrow \infty$ our $s_n \rightarrow L$

My overall plan is to try and use the shrinking of my epsilon to force the sequence to converge. I think this is what to do because since I have to show that the sequence converges this means that all $\epsilon$ are possibilities.

Is my formal proof correct? I really feel that there are some things that I am leaving out to make it airtight.


You can make it easier. Let $s_n$ be an element of $S$ that satisfies $L\leq s_n< L+\frac{1}{n}$. By the squeeze theorem you will get that $s_n$ converges to $L$.

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  • $\begingroup$ See this was my original idea, but my concern was that since $\epsilon$ could be anything that would necessarily mean that I can have an $\epsilon$ smaller than $\frac{1}{n}$ $\endgroup$ – dc3rd Sep 26 '18 at 21:31
  • $\begingroup$ For any $\epsilon>0$ there exists $n\in\mathbb{N}$ such that $\frac{1}{n}<\epsilon$. So with the sequence $\frac{1}{n}$ you can get as close to zero as you wish. $\endgroup$ – Mark Sep 26 '18 at 21:34

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