Closure
(Definition 3.2.11) Given a set \(A \subseteq \mathbf{R}\), let \(L\) be the set of all limit points of \(A\). The closure of \(A\) is defined to be \(\overline{A} = A \cup L\)
Example
Let \(A\) be \(\{1/n: n \in \mathbf{N}\}\). We know that a point \(x=1/n\) is a limit point of a set \(A\) if every \(\epsilon\)-neighborhood \(V_{\epsilon}(1/n)\) of \(1/n\) intersects \(A\) at some point other than \(x\). Set \(\epsilon = 1/n - 1/(n + 1)\). Then
$$
\begin{align*}
V_{\epsilon}(1/n) \cap A = \{\frac{1}{n}\}.
\end{align*}
$$
So this means that \(1/n\) is not a limit point and is an isolated point instead. The only limit point of \(A\) is in fact 0 and so \(\overline{A} = A \cup \{0\}\).
Other Definitions and Properties
- Open Sets
- Limit Points
- Closed Sets
- Closure
- Absolute Value Function
- Sequences, Subsequences and Convergence
- Series and Series Convergence
- Show the Limit Template
References: