Compact Subspace of Linearly Ordered Space

From ProofWiki
Jump to navigation Jump to search


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 proof
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 {{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

Retrieved from "https://proofwiki.org/w/index.php?title=Compact_Subspace_of_Linearly_Ordered_Space&oldid=525187"