Cantor's nested interval theorem is satisfied on every linear continua?

Question

Cantor's nested interval theorem is satisfied on the real line in usual topology. Linear continuum is a generalization of the real line for a topological space. Will this theorem be satisfied on any linear continua?

Answer 1

Yes, it is.

Viewed as a topological space, a linear continuum is a dense linear order $\langle X,\le\rangle$ in which each set bounded above has a supremum, endowed with the order topology. Let $\mathscr{I}=\{I_n:n\in\Bbb N\}$ be a nested family of closed intervals in the linear continuum $X$, where $I_n=[a_n,b_n]$, and $I_n\supseteq I_{n+1}$ for each $n\in\Bbb N$. Then $a_n\le a_{n+1}\le b_{n+1}\le b_n$ for each $n\in\Bbb N$. In particular, $\{a_n:n\in\Bbb N\}$ is bounded above by $b_0$, so it has a supremum $a$. Fix $n\in\Bbb N$; for each $k\ge n$ we have $a_k\le b_k\le b_n$, so $a\le b_n$. Thus, $a\le b_n$ for each $n\in\Bbb N$, and therefore $a\in[a_n,b_n]=I_n$ for each $n\in\Bbb N$, i.e., $a\in\bigcap\mathscr{I}$. This shows that $\bigcap\mathscr{I}\ne\varnothing$.