Let $B \subset \mathbb{R}$ be a non-empty subset. Let $d = inf B$. For every $\epsilon > 0$, the set $[d, d + \epsilon) \cap B$ is non-empty.

Question

I tried to prove this statement by contradiction and say that $\exists \epsilon > 0$ where the set $[d, d + \epsilon) \cap B$ is empty. Because $d \in B$ and $d \in[d, d+\epsilon)$ for every possible $d \in \mathbb{R}$, $d \in [d, d + \epsilon) \cap B$ and the intersection is never empty, therefore a contradiction. I'm not sure if this is sufficient enough to establish the contradiction, because I never used any properties of $d=infraB$ except that it implies $d \in B$.

Answer 1

I'll assume that that $\inf B\ne -\infty$, since the problem makes no sense otherwise. You state that if $d=\inf B$, then $d\in B$, which is not true. This is one of the main differences between $\inf$ and $\min$.

Since $d=\inf B$, $d$ is the largest lower bound of $B$. Now suppose, to the contrary, that $[d,d+\epsilon)\cap B=\emptyset$. Let $d'$ be midpoint of $d$ and $d+\epsilon$; that is, $d'=(2d+\epsilon)/2$. Clearly, $d'>d$. Can you show that $d'$ also a lower bound of $B$, contradicting $d$ being the largest lower bound?