# Compact Subspace of Linearly Ordered Space

This article needs proofreading.Please check it for mathematical errors.If you believe there are none, please remove `{{Proofread}}` from the code.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Proofread}}` from the code. |

This article needs to be linked to other articles.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{MissingLinks}}` from the code. |

## Theorem

Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.

Let $Y \subseteq X$ be a non-empty subset of $X$.

Then $Y$ is a compact subspace of $\struct {X, \tau}$ if and only if both of the following hold:

- $(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
- $(2): \quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$.

## Proof

### Forward Implication

Let $S$ be a non-empty subset of $Y$.

By Compact Subspace of Linearly Ordered Space: Lemma 1, $S$ has a supremum $k$ in $Y$.

This article, or a section of it, needs explaining.In particular: The lemma shows that the structure in question is a complete lattice. Take that further step to assert the existence of the supremum.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

This article, or a section of it, needs explaining.In particular: supremum WRT $\preceq \restriction_Y$You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

Aiming for a contradiction, suppose that $S$ has an upper bound $b$ in $X$ such that $b \prec k$.

Let:

- $\AA = \set {s^\preceq: s \in S} \cup \set {b^\succeq}$

where:

- $s^\preceq$ denotes the lower closure of $s$ in $S$
- $b^\succeq$ denotes the upper closure of $s$ in $S$.

Then $\AA$ is an open cover of $Y$.

But $\AA$ has no finite subcover, contradicting the fact that $Y$ is compact.

This needs considerable tedious hard slog to complete it.In particular: above lines need proofTo discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

This article, or a section of it, needs explaining.In particular: Pronounce proper incantations about duality.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

A similar argument proves the corresponding statement for infima, so $(2)$ holds.

$\Box$

### Reverse Implication

Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.

Let $Y \subseteq X$ be a non-empty subset of $X$.

Let the following hold:

- $(1): \quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
- $(2): \quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$.

Then $Y$ is a compact subspace of $\struct {X, \tau}$.

## Also see

## Sources

- 1955: John L. Kelley:
*General Topology*: $\S 5$: Problem $\text C$