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The truth is, I don't even know where to begin with this question... Can someone please offer their assistance?

Vladimir Nabokov
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  • The first hint really says everything. You know that for each $n$ there is an $x_n \in S$ which satisfies $x_n < \inf S + \frac1n$ (by definition of the infemum). Show that this sequence converges to $\inf S$. Edit: Also, note that $\inf S < x_n < \inf S + \frac1n$. – Tyler Holden Oct 23 '13 at 21:01

2 Answers2

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Let $n\in\mathbb N$. By definition of $\inf S$, the number $\inf S+\frac 1n$ is not a lower bound of $S$, hence there exists an element $x_n\in S$ with $x_n<\inf S+\frac 1n$. On the other hand $x_n\ge \inf S$ because $\inf S$ is a lower bound. We conclude that $|x_n-x|<\frac 1n$ and hence $x_n\to x$.

Hagen von Eitzen
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By the definition of the infimum $\inf S$ we have $$\forall \epsilon>0\quad \exists x\in S\ \text{such that}\ \inf S\leq x<\inf S+\epsilon$$ so take $\epsilon=\frac 1 n, \ n\in\mathbb{N}^*$ we find $x_n\in S$ such that $\inf S\leq x_n<\inf S+\frac 1 n$ and passing to the limit in the last inequalities gives $\displaystyle\lim_{n\to\infty}x_n=\inf S$.