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

(1)any $s \in S$, $t \in T$$s<t$

(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$, $\inf T<t$, $s>t$,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?

  • $\begingroup$ 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? $\endgroup$
    – AMC3F9
    Feb 12 at 3:31
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 12 at 4:06


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