# Is $K:=\{\frac{1}{n}\mid n \in N\} \cup \{0\}$ compact in $T_{st}$?

this is a topology question:

Is $K:=\{\frac{1}{n}\mid n \in N\} \cup \{0\}$ compact in $T_{st}$?

My intuition: it is now compact, because $0$ is already covered, and also I proved before that without the $0$, $K$ is not compact in standard topology because we can never cover $0$.

But my question is, now we include $0$, but isn't the cardinality of $K$ still infinite considering the natural number is infinite? So does this mean we still can't find a finite sub-covering for $K \cup \{0\}$ under standard topology?

Thank you in advance for the help!

Standard topology:=The collection $B_{st} := \{(a, b)\}$ defines a basis for a topology on $\mathbb{R}$, called the standard topology.

• What is $T_{st}$? If it means "Standard Topology" I think it doesn't make sense to wonder whether $T_{st}$ is compact in $K$. The natural question would be wheter $K$ is compact in $T_{st}$. – Javi Mar 1 '18 at 22:16
• @Javi I was typing in a hurry, you are correct and I just edit the question, thanks for catching this dump mistake! – Liz Mar 1 '18 at 22:20
• $K$ minus $0$ is not compact because it is not closed. Or because we can find an open cover of it that does not have a finite subcover. Not "because we cannot cover $0$", we don't even need to cover a poitn not in the set.. – Henno Brandsma Mar 1 '18 at 22:20

The correct formulation is: is $K = \{0\} \cup \{\frac{1}{n}: n \in \mathbb{N}^+\}$ compact with respect to $\mathcal{T}_{\text{st}}$? (a set is compact in a topology, not a topology is compact in a set.)
The answer is yes: Let $\{(a_i, b_i): i \in I\}$ be any set of basic open intervals of $\mathbb{R}$ such that $K \subseteq \cup_{i \in I} (a_i, b_i)$. Because $0 \in K$ we must have some $(a_{i_0}, b_{i_0})$ that contains $0$.
Now $a_{i_0} < 0$ and $b_{i_0} > 0$ (or $0$ would not be in this interval) and by standard properties of the reals (Archimedean property) we have that for some $n_0 \in \mathbb{N}^+$ we have that $\frac{1}{n_0} < b_{i_0}$. This implies that $a_{i_0} < 0 < \frac{1}{n} \le \frac{1}{n_0} < b_{i_0}$ for all $n \ge n_0$ and thus that the set $(a_{i_0}, b_{i_0})$ also already contains all $\frac{1}{n}$ for $n \ge n_0$ and also $0$. So all points of $K$ are covered by this one set, except possibly the finitely many points $\frac{1}{1},\ldots,\frac{1}{n_0 -1}$.
So picking an open interval $(a_{i_j}, b_{i_j})$ to cover $\frac{1}{j}$ as well for these finitely many $j$ we have a finite subcover of the original cover and so we can find a finite subcover for every such open cover, and $K$ is indeed compact.
In $T_{st}$ you've got a characterization for compact sets, i.e., a set is compact in $T_{st}$ if and only if it is closed and bounded. $K$ is clearly bounded because it is cointained in $[0,1]$ and it is closed because it cointain all its limit points (it is easy to check using the general definition of limit point).