Deleted Integer Topology is Weakly Countably Compact

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $S = \R_{\ge 0} \setminus \Z$.

Let $\tau$ be the deleted integer topology on $S$.


Then the topological space $T = \left({S, \tau}\right)$ is weakly countably compact.


Proof

Let $A \subseteq S$ such that $A$ is infinite.

Let $x \in A$.

Let $n \in \Z$ such that $n < x < n + 1$.

By the definition of the deleted integer topology, $\left({n \,.\,.\, n + 1}\right)$ is open in $T$.


We have that $\left({n \,.\,.\, n + 1}\right)$ is infinite.

Take some $y \in \left({n \,.\,.\, n + 1}\right)$ such that $x \ne y$.

Now we claim that $y$ is a limit point of $A$.


Let $N_y$ be an open neighbourhood of $y$.

To be open in $T$, $N_y$ needs to be the union of open intervals of the form $\left({k \,.\,.\, k + 1}\right)$ for some $k \in \Z_{\ge 0}$.

Therefore to contain $y$ and be open at the same time, $\left({n \,.\,.\, n + 1}\right) \subseteq N_y$.

Thus $x \in N_y$ and $y \ne x \in A \cap N_y \ne \varnothing$.

This is the definition of a limit point.

So, by definition, $T$ is weakly countably compact.

$\blacksquare$


Sources