A Compact Set that is not Closed

Last updated Nov 1, 2022

# Counterexample

Suppose $\mathbb{Z}$ is given the Cofinite Topology. Suppose $S \subset \mathbb{Z}$. Then $S$ is Compact. In particular there exists, $S \subset \mathbb{Z}$ that is Compact but not Closed.

Let $\mathcal{U} = {U_{\alpha}}{\alpha \in I}$ be an Open Cover for $S$. By definition of $\mathbb{Z}$, $|U{\alpha}^{C}| < \infty$ for each $\alpha \in I$. Let $\alpha_{0} \in I$. Let $U_{\alpha_{0}}^{C} \cap S =: {x_{i}}{i=1}^{n}$ for $n = |U{\alpha_{0}}^{C} \cap S| < \infty$. Since $\mathcal{U}$ is an Open Cover of $S$, for each $i \in [n]$, there exists an $\alpha_{i} \in I$ so that $U_{\alpha_{i}} \ni x_{i}$.Thus, ${U_{\alpha_{0}}, \dots, U_{\alpha_{n}}}$ is a finite Open Subcover for $S$ and $S$ is Compact.

Simply finding an $S \subset \mathbb{Z}$ that is not Closed will prove the second assertion. Recall $K \subset \mathbb{Z}$ is Closed If and Only If there is an Open $U \subset \mathbb{Z}$ s.t. $U^{C} = K$. Since $\mathbb{Z}$ has the Cofinite Topology, $K$ must be a Finite Set or $\mathbb{Z}$ itself. So $\mathbb{Z} \setminus {0}$ is not Closed, but by above, it is Compact. $\blacksquare$