Let $\beta \nat$ be the set of all [[Ultrafilter|ultrafilters]] on $\nat$. For each $\emptyset \ne A \subset \nat$, let $U_A = \bracs{\fU \in \beta \nat: A \in \fU}$ be the set of ultrafilters containing it, then 1. The set $\ce = \bracs{U_A: \emptyset \ne A \subset \nat}$ is a [[Base|base]] for a [[Topological Space|topological space]]. 2. $\beta \nat$ is a [[Stone-Čech Compactification]] of $\nat$. *Proof*. $(1)$: Every [[Filter|filter]] on $\nat$ belongs to $U_\nat$. If $\emptyset \ne A, B \subset \nat$ with $A \cap B \ne \emptyset$, then $U_{A} \cap U_B = U_{A \cap B}$. $(2)$: Firstly, there is an injection $\nat \to \beta \nat$ with $n \mapsto \wh n = \fF(\bracs{n})$. Since each $\bracsn{\wh n} = U_{\bracs{n}}$, the subspace topology is discrete. For each $\emptyset \ne A \subset \nat$, there exists $n \in \nat$ such that $n \in A$ and $\wh n \in U_{\bracs{n}} \subset U_A$. Therefore $\overline{\nat} = \beta \nat$. Let $\fU_1, \fU_2 \in \beta \nat$ with $\fU_1 \ne \fU_2$, then there exists $A \subset \nat$ such that $A \in \fU_1$ but $A \not\in \fU_2$. In which case, $\fU_1 \in U_A$ and $\fU_2 \in U_{A^c}$ with $U_A \cap U_{A^c} = \emptyset$. Let $\bracsn{U_{A_i}}_{i \in I}$ be an [[Open Cover|open cover]] of $\beta \nat$ with no finite subcover. Let $\fB = \bracs{A_i^c: i \in I}$, then since $\bracs{U_{A_i}}_{i \in I}$ has no finite subcover, there exists no $\bracsn{A_{i_j}}_1^n$ such that $\bigcup_{j = 1}^nA_{i_j} = \nat$ and $\bigcap_{j = 1}^n A_{i_j}^c = \emptyset$. Otherwise, for each $\fU \in \beta \nat$, there exists $1 \le j \le n$ such that $A_{i_j} \in \fU$. Therefore $\fB$ is a filter base. Let $\fU$ be an ultrafilter containing $\fB$, then $\fU$ does not contain $A_i$ for all $i \in I$. Hence $\bracsn{U_{A_i}}_{i \in I}$ is not a cover. Define as accumulation points $ \bracs{n \in \nat: |L - f(n)| < \eps} \in \fU $