I have this question here:
Let $f\colon \mathbb{R} \rightarrow \mathbb{R}$ and let $m \in \mathbb{R}$. Consider the set:
$A = \{x \in [a,b) \mid f(x) > m\}$, $A$ is not empty.
Prove that $s=\inf(A)$ exists and that $a\leq s<b$.
Can I say that $A$ is bounded from below by $a$? What is the explanation for that? Is this because $a\leq x<b$? But what about $f(x)$?
From here, should I use $\varepsilon$ somehow?
Thank you for your help and for your time!
On
Since $a$ is a lower bound of $A$ (i.e. $a\leq x$ for all $x\in A$) we need to show that it is indeed the greatest lower bound of $A$ (this is the definition of an infimum). Suppose for contradiction there exists an element $c>a$ that is also a lower bound of $A$. But $a\in A$ and by assumption $a<c$, hence $c$ can’t be a lower bound on $A$. Therefore $\inf(A)$ exists, namely $a$.
Next we need to show that $a\leq a < b$. Clearly this boils down to showing that $a<b$. Note that $b$ is an upper bound of $A$ and also $b\notin A$. This implies $x<b$ for all $x\in A$. In particular, $a<b$, as required.
Your set $A$ is, by definition, a subset of $[a,b)$. So, $a$ is a lower bound of $A$ and therefore, since $\inf A$ is the greatest lower bound of $A$, $a\leqslant\inf A$.
And there is some $c\in A$. Since $c\in[a,b)$, $c<b$. So, $\inf A\leqslant c<b$.