# Is this a Legitimate Proof Regarding the Infimum of a Set?

The statement reads as follows:

If $$S \subset \mathbb R$$ is bounded below and $$w=inf S$$, show that for any $$n \in \mathbb N$$ there exists $$x_n \in S$$ with $$w \le x_n \lt w + {\frac 1 n}$$

I came up with something like this:

1. By the definition of infimum we know that $$\forall x \in S, w \le x$$. So $$w$$ is a lower bound for the set $$S$$.

2. By the Archimedean Property, $$\forall \epsilon > 0, \exists n \in \mathbb N$$ s.t. $${\frac 1n} \lt \epsilon$$. Which gives $$w+\epsilon \lt w+ {\frac 1n}$$. And since $$w+\epsilon$$ clearly isn't an upper bound for S then $$w \le x_n \lt w + {\frac 1 n}$$.

I'm mostly concerned about the second part of that proof, I'm not 100% sure it proves the above statement above...

Also side question: in the Epsilon definition of Supremum/Infimum (Let S be a nonempty subset of the real numbers that is bounded above. The upper bound u is said to be the supremum of S if and only if $$∀ϵ>0$$ there exists an element $$x_ϵ∈S$$ such that $$u−ϵ) what's the importance of the $$\epsilon$$ in $$x_\epsilon$$? Other than signifying some arbitrary number on the real number line.

• You do not need Archimedean, just note directly taht $w+\frac 1n>w$, hence $w+\frac1n$ cannot be a lower bound. – Hagen von Eitzen Oct 5 '18 at 22:32

The second part is simple than you think. If $$n\in\mathbb N$$, then $$\frac1n>0$$ and therefore there is a $$x_n\in S$$ such that $$w\leqslant x_n. There is no need to introduce a $$\varepsilon$$ here.
• @clovis It is not wrong, but i is hard to understand what that $\varepsilon$ is doing there… – José Carlos Santos Oct 5 '18 at 22:42
By definition, $$w$$ is the greater lower bound of $$S$$ and so the proposition follows. Do you understand why?