Approximation Property for Infimum

Question

Could someone tell me whether I wrote the approximation property for infimum correctly and whether the proof is okay too? I based this on Apostal.

Let $S$ be a nonempty set of real numbers with an infimum $b$. Then for any real number $a>b$, there exists some real number $x$ in $S$ such that $b\leq x<a$.

The part $b\leq x$ is true by the definition of lower bounds. Assume that $a\leq x$ for all $x$ in $S$. Then $a$ is a lower bound for $S$. This forms a contradiction ($a$ is a lower bound for $S$ and $a>b$ yet $b$ is the greatest lower bound for $S$) by assuming $a\leq x$ for all $x$ in $S$.

Answer 1

Yes, you've correctly written the approximation property for infimum and your proof is also correct. Note that we need $S$ to be bounded below to have an infimum so you may want to include this condition of $S$ to the property.

Another way of writing the approximation property (using notation $\varepsilon$) is that:

If $b$ is the infimum of nonempty set $S$ then for any $\varepsilon>0$, there exists $a \in S$ such that $b \le a<b+\varepsilon$.

Answer 2

Yes, this follows from the $\inf, \sup$ being a limit point (there's a sequence that converges to them).

Let $a$ be the inf of $S$, which is non-empty and bounded below(I'll leave the unbounded as an exercise). Then suppose there is no sequence $\{a_k\} \subset S$ such that $a_k \rightarrow a$.

Fix $x \in [a,a+1)$. Indeed, we must first show $[a,a+1)$ is non-empty. Suppose it is empty, then we see that as $a$ is a lower bound, there is no $x \in (-\infty,a) \cap S$. However, notice as $(a,a+1)$ is empty, we must have that $a+1$ is another lower bound; however, $a+1 > a$, but $a$ is the greatest lower bound(inf), so we have a contradiction. Therefore, $[a,a+1)$ is non-empty. So let $a_1 = x$.

Now repeat the proof above for $a_n$ where $a_n \in [a,a+1/n)$. Then we see as $1/n \rightarrow 0$ as $n \rightarrow \infty$, $a_n \rightarrow a$. Hence, by definition of convergence, for any $\epsilon > 0$, there is some $N \in \mathbb{N}$ such that $|a-a_n| < \epsilon$, hence by the definition of inf, we have $a < \epsilon + a_n$. where $a_n \in S$.

This gives your result, and proves the other answers statement. A symmetric results also holds for the sup of a sequence.