Deleted Integer Topology is not Countably Compact
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Theorem
Let $S = \R_{\ge 0} \setminus \Z$, and let $\tau$ be the deleted integer topology on $S$.
Then the topological space $T = \struct {S, \tau}$ is not countably compact.
Proof
By definition, the deleted integer topology is a partition topology.
Let $\PP$ be the partition which is the basis for $T$, that is:
- $\PP = \set {\openint {n - 1} n: n \in \Z_{> 0} }$
Then $\PP$ is a countable open cover of $S$ which has no finite subcover.
Hence the result.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $7$. Deleted Integer Topology: $5$