# 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?

I look at the answer of another post, but still have the above confusion in my head about the finiteness. Please help clear my confusion.

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

## 2 Answers

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).