6

Every finite set of reals has the property that every non-empty subset of it has a maximum and a minimum. Does this property characterize finite sets of reals? That is, is it the case that if $S$ is a subset of the real numbers where every non-empty subset of $S$ has a maximum and a minimum, then $S$ is finite?

user107952
  • 20,508
  • 1
    @RahulMadhavan, $[0,\frac12)$ is a non-empty subset of $[0,1]$ without a maximum. – Martund Apr 08 '21 at 11:38
  • Yes, realized and retracted my statement. Thanks for pointing out. @Martund – Rahul Madhavan Apr 08 '21 at 11:40
  • @user3733558, $\mathbb Q\cap [0,\frac12)$ is a non-empty subset of that without a maximum, again. – Martund Apr 08 '21 at 11:42
  • Good question! I has not thought about this earlier. – Kavi Rama Murthy Apr 08 '21 at 11:45
  • This has nothing to do with the real numbers - for any linear order $(S,\leq)$, $S$ is finite if and only if every subset of $S$ has a maximum and a minimum. See this question: https://math.stackexchange.com/q/3504285/7062. The topological component of @KaviRamaMurthy's nice answer (every bounded sequence in $\mathbb{R}$ has a limit point) can be replaced by an application of Ramsey's theorem as in Henno Brandsma's answer or by the slicker argument in David C. Ullrich's answer. – Alex Kruckman Apr 16 '21 at 15:17

1 Answers1

5

True. Suppose $S$ is an infinte set. If it is unbounded it is clear that there is a subset with no maximum or a subset no minimum. Suppose it is bounded. Let $(x_n)$ be sequence of distinct points of $S$. Then either there is a subsequence which increases to some limit $x$ or there is a subsequence which decreases to some limit $x$. In the first case $(x_{n_k})$ has no maximum and in the second case $(x_{n_k})$ has no minimum.

Kavi Rama Murthy
  • 311,013