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

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Theorem

Let $X = \left[{0 \,.\,.\, 1}\right) \cup \left({2 \,.\,.\, 3}\right) \cup \left\{{4}\right\}$.

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 $Y = \left[{0 \,.\,.\, 1}\right) \cup \left\{{4}\right\}$.

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


Then:

$\left({Y, \preceq}\right)$ is a complete lattice
$Y$ is closed in $X$

but:

$\left({Y, \tau'}\right)$ is not compact.


Proof

First it is demonstrated that $\left({Y, \preceq}\right)$ is a complete lattice.

Let $\phi: Y \to \left[{0 \,.\,.\, 1}\right]$ be defined as:

$\phi \left({y}\right) = \begin{cases} y & : y \in \left[{0 \,.\,.\, 1}\right) \\ 1 & : y = 4 \end{cases}$

Then $\phi$ is a order isomorphism.



$\Box$

We have that $\left[{0 \,.\,.\, 1}\right]$ is a complete lattice.



Next is is shown that $\left({Y, \preceq}\right)$ is closed in $X$.

Let $x \in X \setminus Y$.

Then:

$x \in \left({2 \,.\,.\, 3}\right)$

Thus:

$x \in \left({\dfrac {x+2} 2 \,.\,.\, \dfrac {x+3} 2}\right) \in \tau'$

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


Finally it is shown that $\left({Y, \tau'}\right)$ is not compact.

Let:

$\mathcal A = \left\{ {x^\preceq: x \in \left[{{0}\,.\,.\,{1}}\right) }\right\} \cup \left\{ {\left({\dfrac 5 2}\right)^\succeq}\right\}$

where:

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

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



$\blacksquare$