Closed Sets
An isolated point always belongs to the set \(A\) unlike a limit point where it’s not necessarily an element of the set!
Proof: here.
Example 1: \(\mathbf{Z}\). Why? The complement of a closed set is open. The complement of \(\mathbf{Z}\) is the union of all the open intervals \(\{...\cup ... (0,1)\cup(1,2)\cup...\}\). But we know that the union of a collection of open sets is open. Therefore, \(\mathbf{Z}\) is closed. And why was it not open? Because the definition of an open set requires an entire neighborhood around each element entirely contained in \(\mathbf{Z}\) but here if you take any integer, the neighborhood around the integer will contain rationals and irrationals.
Example 2: \(\{\frac{1}{n}: n \in \mathbf{N}\} \cup \{0\}\). This set is not open since we won’t be able to find a neighborhood around each element where it’s entirely contained in the set. This set is closed since it contains the only limit point which is 0.
Example 3: \(\{\mathbf{R}\} \cup \{0\}\). This set is open since we can find a neighborhood around each element contained in \(\mathbf{R}\). This set is also closed since it contains all of its limit points!
Proof: TODO.
Other Definitions and Properties
- Open Sets
- Limit Points
- Compact Sets
- Closure
- Absolute Value Function
- Sequences, Subsequences and Convergence
- Series and Series Convergence
- Show the Limit Template
References: