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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 $u−ϵ<x_ϵ$) what's the importance of the $\epsilon$ in $x_\epsilon$? Other than signifying some arbitrary number on the real number line.

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    $\begingroup$ You do not need Archimedean, just note directly taht $w+\frac 1n>w$, hence $w+\frac1n $ cannot be a lower bound. $\endgroup$ Oct 5, 2018 at 22:32

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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 $w\leqslant x_n<w+\frac1n$. There is no need to introduce a $\varepsilon$ here.

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  • $\begingroup$ Ah that's true. So would my proof be wrong, or just overkill but still right? $\endgroup$
    – clovis
    Oct 5, 2018 at 22:35
  • $\begingroup$ @clovis It is not wrong, but i is hard to understand what that $\varepsilon$ is doing there… $\endgroup$ Oct 5, 2018 at 22:42
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This is directly by the definition of infimum and nothing more.

By definition, $w$ is the greater lower bound of $S$ and so the proposition follows. Do you understand why?

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