# 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 $L$

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 $$L$$

Attempt

($$\Longrightarrow$$) 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 $$L$$.
• 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 $\frac{1}{n}$ – dc3rd Sep 26 '18 at 21:31
• 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. – Mark Sep 26 '18 at 21:34