(Definition 3.2.6) A point \(a \in A\) is an isolated point of \(A\) if it is not a limit point of \(A\).

An isolated point always belongs to the set \(A\) unlike a limit point where it’s not necessarily an element of the set!

(Definition 3.2.7) A set \(F \subseteq \mathbb{R}\) is closed if it contains its limit points.

Proof: here.
Example 1: \(\mathbb{Z}\). Why? The complement of a closed set is open. The complement of \(\mathbb{Z}\) 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, \(\mathbb{Z}\) 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 \(\mathbb{Z}\) but here if you take any integer, the neighborhood around the integer will contain rationals and irrationals.
Example 2: \(\{\frac{1}{n}: n \in \mathbb{N}\} \cup \{0\}\). 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: \(\{\mathbb{R}\} \cup \{0\}\). This set is open since we can find a neighborhood around each element contained in \(\mathbb{R}\). This set is also closed since it contains all of its limit points!

(Theorem 3.2.8) A set \(F \subseteq \mathbb{R}\) is closed if and only if every Cauchy sequence contained in \(F\) has a limit that is also an element of \(F\).

Proof: TODO.
Other Definitions and Properties


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