# Prove that $\sup S = \inf T$

Let both $$S$$ and $$T$$ be non-empty subsets of $$\Bbb{R}$$ and satisfy

（1）any $$s \in S$$, $$t \in T$$$$s

（2）any $$\varepsilon > 0$$, $$\exists s_0 \in S$$, $$t_0 \in T$$$$t_0 - s_0 < \varepsilon$$

Show that $$\sup S = \inf T$$.

I tried to proof by contradiction，$$\sup S ＞ \inf T$$$$\sup S ＜ \inf T$$， when $$\sup S > \inf T$$$$\sup S>s$$, $$\inf T, $$s>t$$，it's false.

When $$\sup S < \inf T$$, $$t_0-s_0<\varepsilon$$, $$t_0 > \inf T$$, $$s_0<\sup S$$, $$t_0-s_0>\inf T - \sup S>0$$

But what do I do next?

• You got that $t_0-s_0>\text{inf T}-\text{sup }S$, but $t_0-s_0<\epsilon$, i.e, $\text{inf }T-\text{sup }S<\epsilon$. Can you take it from there? Feb 12 at 3:31
• Your first case has issues. Putting your inequalities, end to end, you get $s < \sup S > \inf T < t$, which tells you nothing about how $s$ and $t$ are related. Instead, I would claim that there exists some $s \in S$ such that $s > (\sup S + \sup T) / 2$, because $(\sup S + \sup T) / 2 < \sup S$ (under the assumption that $\sup S > \inf T$), otherwise $(\sup S + \sup T) / 2$ is a lesser upper bound on $S$. Similarly, there exists $t \in T$ such that $t < (\sup S + \sup T) / 2$, otherwise $(\sup S + \sup T) / 2$ is a greater lower bound on $T$. From there, $s > t$, as required. Feb 12 at 4:06