> [!quote] Idea > > Dense subsets cannot be separated from the space with open sets. Every element can be approximated with elements from a dense subset via nets. If the space is first countable, with sequences. > [!definition] > > Let $(X, \topo)$ be a [[Topological Space|topological space]]. A [[Set|set]] $A \subseteq B \subseteq X$ is **dense** in $B$ if the [[Topological Closure|topological closure]] $\overline{A \cap B} = \ol{B}$. > [!theorem] > > The following characterisations of a set $A$ being dense in $X$ are equivalent > - $\ol{A} = X$ > - Every non-empty [[Open Set|open set]] intersects $A$. > - $A$ is dense in every open subset of $X$. > > *Proof*. Suppose that $\ol{A} = X$, and let $U$ be a non-empty open set. Since any points $x \in U$ adheres to $A$, $U \cap A$ is non-empty. Suppose that every non-empty open set intersects $A$, then any point is an adherent point of $A$. Therefore $\ol{A} = X$. > > Suppose that $\ol{A \cap U} = \ol{U}$ for any open set $U$, then $\ol{A} = \ol{X} = X$ and $A$ is dense in $X$. Suppose that every non-empty open set intersects $A$, and let $U$ be an open set. Then for any $x \in \ol{U}$, we have $V \cap U \ne \emptyset \forall V \in \cn(x)^c$. Since $A$ intersects every non-empty open set, $V \cap U \cap A \ne \emptyset \forall V \in \cn(x)^c$ and $x \in \ol{U \cap A}$. Therefore $\ol{A \cap U} = \ol{U}$. > [!definition] > > A set $A \subseteq X$ is *nowhere dense* if the [[Interior|interior]] $\inte A = \emptyset$. > [!definition] > > A set $B \subseteq X$ is **separable** if $\exists A \subseteq B$ where $A$ is [[Cardinality|countable]] and dense in $B$.