Subset of Linearly Ordered Space which is Order-Complete and Closed but not Compact

Theorem

Let $X = \hointr 0 1 \cup \openint 2 3 \cup \set 4$.

Let $\preceq$ be the ordering on $X$ induced by the usual ordering of the real numbers.

Let $\tau$ be the $\preceq$ order topology on $X$.

Let $Y = \hointr 0 1 \cup \set 4$.

Let $\tau'$ be the $\tau$-relative subspace topology on $Y$.

Then:

$\struct {Y, \preceq}$ is a complete lattice

$Y$ is closed in $X$

but:

$\struct {Y, \tau'}$ is not compact.

Proof

First it is demonstrated that $\struct {Y, \preceq}$ is a complete lattice.

Let $\phi: Y \to \closedint 0 1$ be defined as:

$\map \phi y = \begin{cases}

y & : y \in \hointr 0 1 \\ 1 & : y = 4 \end{cases}$

Then $\phi$ is a order isomorphism.

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$\Box$

We have that $\closedint 0 1$ is a complete lattice.

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Next is is shown that $\struct {Y, \preceq}$ is closed in $X$.

Let $x \in X \setminus Y$.

Then:

$x \in \openint 2 3$

Thus:

$x \in \openint {\dfrac {x + 2} 2} {\dfrac {x + 3} 2} \in \tau'$

Since the complement of $Y$ is open, $Y$ is closed.

Finally it is shown that $\struct {Y, \tau'}$ is not compact.

Let:

$\AA = \set {x^\preceq: x \in \hointr 0 1} \cup \set {\paren {\dfrac 5 2}^\succeq}$

where:

$x^\preceq$ denotes the lower closure of $x$

$x^\succeq$ denotes the upper closure of $x$

Then $\AA$ is an open cover of $Y$ with no finite subcover.

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$\blacksquare$

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