Let $X$ be a totally ordered set and $\forall U\subseteq X$ s.t. $U\neq\emptyset$ has both a max. and min., then $X$ is finite.

Question

Proof: Let $X$ be a totally ordered set and $\forall U\subseteq X$ s.t. $U\neq\emptyset$ has both a minimum and a maximum. We want to show that X is finite. For the sake of contradiction, suppose $X$ is infinite. We should show $\exists U\subseteq X$ where $U $does not have both a minimum and a maximum. Which, equivalently, mean that $\forall U\subseteq X$ such that $U$ has both a max. and a min. is false. Now we define $X_0:=\{\alpha\in X:x_0\leq\alpha\leq x_n\}$ where $x_0$ is the minimum and $x_n$ is the maximum. In case that, $\forall x\in X, x\in X_0$ then $X\subseteq X_0$ and $X=X_0$. So we consider $X^*\subseteq X_0$ such that $X^*:=\{\alpha_j\in X_0:x_0\leq\alpha_j< x_n \}=X_0\setminus \{x_n\}$. Since $\forall \alpha_k\in X^*, \alpha_k\leq x_n$ and $X^*\subset X$ then $X^*$ is the subset of $X$ not containing both a maximum and a minimum which yields a contradiction.

Answer 1

Your proof is doomed to fail. There are infinite linear orders where every non-empty subset has a minimum. It's just that not every subset has a maximum. This is the case with $\Bbb N$ with the usual order. And there are other linear orders with the converse property.

The point is that the negation of "Every non-empty subset has a minimum and maximum" is "There is a non-empty subset without a minimum or without a maximum".

So your goal needs to be to assume that $X$ is an infinite linear order where every set has a minimum, and find a subset which does not have a maximum. Or assume that every subset has a maximum, and find a subset without a minimum element.

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But in any case, a direct proof is also possible. Let $U$ be the set of points that have infinitely many points below them in $X$. If $U$ is non-empty, it has a minimum, $x$ and you can show that $\{y\in X\mid y<x\}$ has no maximum, since every $y$ in that set has finitely many predecessors, but $x$ has infinitely many of them. So $U$ has to be empty. Apply a similar argument to tail segments to show that there is no point in $X$ that has infinitely many points above it.

Now pick any point $x$, since $X$ is a union of those points below $x$, those points above $x$, and $\{x\}$, and all three sets are finite, $X$ is finite.

Answer 2

Assume X is infinite and let $x_1$ be the bottom element of X.
Let $X_1$ = X - {$x_1$} and $x_2$ the bottom element of $X_1$.
Continue in this manner creating the set { $x_j$ : j in N }.
which has no maximum.