> [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]] and $A \subseteq X$, the **topological closure** of $A$, > $ > \ol{A} = \bigcap_{B^c \in \topo, B \supset A}B > $ > is the intersection of all [[Closed Set|closed sets]] containing $A$. > [!definition] > > Let $(X, \topo)$ be a topological space and let $A \subset X$. A point $x \in X$ is an **adherent point** of $A$ if every [[Neighbourhood|neighbourhood]] of $x$ intersects $A$/no neighbourhood of $x$ separates $x$ from$A$. > [!theoremb] Theorem > > Let $(X, \topo)$ be a topological space, $A \subset X$, $x \in X$ and $\cm(x)$ be a [[Neighbourhood Base|neighbourhood base]] at $x$. Then the following are equivalent: > - $x \in \ol{A}$. > - $x$ adheres to $A$. > - No neighbourhood $U \in \cm(x)$ separates $x$ from $A$. > - There exists a [[Net|net]] $\angles{x_\alpha}_{\alpha \in \cm(x)} \subset A$ such that $x_\alpha \to x$. > - *(If $x$ has a countable neighbourhood base)* There exists a [[Sequence|sequence]] $\seq{x_j}$ that [[Limit|converges]] to $x$. > > *Proof*. Firstly, > $ > x \in \ol{A}^c \Leftrightarrow \exists U \in \cn(x)^o: U \cap A = \emptyset \Leftrightarrow x \text{ doesn't adhere to }A > $ > Secondly, if $x$ adheres to $A$, then no open neighbourhoods separate $x$ from $A$, including the neighbourhood base. Suppose that no neighbourhood $U \in \cm(x)$ separates $x$ from $A$. Then for any neighbourhood $V \in \cn(x)^o$, there exists $U \in \cm(x)$ such that $x \in U \subset V$ and $U \cap A \ne \emptyset$ ($V \cap A \ne \emptyset$) and $V$ does not separate $x$ from $A$. > > Thirdly, suppose that $x$ adheres to $A$, then we can choose $x_{\alpha} \in A \cap \alpha$ for all $\alpha \in \cm(x)$. Then for any $V \in \cn(x)^o$, there exists $\beta \in \cm(x)$ such that $x_\alpha \in \alpha \subset \beta \subset V$ for all $\alpha \gtrsim \beta$, and $x_\alpha \to x$. If $x$ does not adhere to $A$, then let $U \in \cm(x)$ such that $U \cap A = \emptyset$. This means that no net in $A$ can converge to $x$. > > Lastly, if $\cm(x)$ is countable neighbourhood base, we can reduce it to $\seq{U_j}$ where $U_j \supset U_{j + 1}$ for all $j \in \nat$. Choose $x_j \in A \cap U_j$, then for any $V \in \cn(x)^o$ there exists $U_k$ such that $x \in U_{j} \subset U_k \subset V$ for all $j \ge k$. Therefore $x_j \to x$. If $x$ does not adhere to $A$, then let $U \in \cm(x)$ such that $U \cap A = \emptyset$. This means that no sequence in $A$ can converge to $x$. > [!definition] > > Let $A \subset X$, and $(Y \subset X, \topo_{Y})$ be an [[Relative Topology|induced topology]]. The topological closure of $A$ in $Y$, $\overline{A}_{Y}$ is > $ > \overline{A}_{Y} = \overline{A} \cap X > $