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Let $S$ and $T$ be subsets of $\mathbb R$ such that $s \lt t$ for each $s \in S$ and each $t \in T$. Prove carefully that $$\sup S \le \inf T$$

Best way to prove such a question?

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  • $\begingroup$ By writing down the definition of supremum and infimum and applying them. Can you share what you've tried and what you're having trouble with? $\endgroup$
    – user296602
    Apr 17 '16 at 6:01
  • $\begingroup$ We should specify that $S$ and $T$ are non-empty. $\endgroup$ Apr 17 '16 at 6:02
  • $\begingroup$ Maybe you could have a look at this post and try whether you can use somewhat similar arguments. $\endgroup$ Apr 17 '16 at 11:36
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Every $t \in T$ is an upper bound for $S$. Thus, $\sup S \leq t$, for all $t \in T$ by definition of the supremum. This inequality implies $\sup S$ is a lower bound for $T$ and so, by definition, $\sup S \leq \inf T$.

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Suppose this is not true. Let $\sup S \gt \inf T$. By the denseness of reals, there exists some $k\in\mathbb R$ such that $$\inf T \lt k \lt \sup S$$

Suppose $k\in S$. Then, for all $x\lt k$, $x\notin T$, since we must have $k$ less than any element of $T$. Then, for all $t\in T$, $k\le t$. Thus, $k$ becomes a lower bound of $T$ greater than $\inf T$, which is a contradiction. Thus, $k\notin S$. Similarly, it can be shown that $k\notin T$.

Thus, for all $t\in T$, $t\notin (\inf T,\sup S)$. But $t\gt \inf T$. Then, we have $\sup S\lt t$ for all $t\in T$. Thus, we have a lower bound of $T$ which is greater than $\inf T$ which is a contradiction.

Thus, the assumption is incorrect and $$\sup S\le\inf T$$

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