----- > [!proposition] Proposition. ([[neighborhood-basis characterization of set closure]]) > Let $A$ be a subset of the [[topological space]] $X$. > 1. Then $x \in \overline{A}$ if and only if every [[neighborhood]] $U$ of $x$ intersects $A$ nontrivially. > 2. Supposing the [[topological space|topology]] of $X$ is given by a [[basis for a topology|basis]], then $x \in \overline{A}$ if and only if every [[basis for a topology|basis element]] $B \ni x$ intersects $A$ nontrivially. > [!proof]- Proof. ([[neighborhood-basis characterization of set closure]]) > -- tried doing this on my own, munkres proof is a bit more straightforward and should be consulted > 1. We will show the contrapositive: $x \notin \overline{A} \iff \text{there exists a neighborhood } U \text{ of $x$ that trivially intersects } A. $ Suppose $x \notin \overline{A}$. Then there exists a [[closed set]] $C \in X$ for which $A \subset C$ but $x \notin C$. Observe then that $X - C$ is [[open set|open in]] $X$, contains $x$, and trivially intersects $A$. \ Conversely, suppose there exists a [[neighborhood]] $U \ni x$ with $U \cap A= \emptyset$. Then $X-U$ is [[closed set|closed in]] $X$ and contains $A$, but does not contain $x$ — as desired. \ >2. This follows readily. Suppose $x \notin \overline{A}$. Then as before we have a [[closed set]] $C \in X$ for which $A \subset C$ but $x \notin C$. $X - C$ is [[open set|open in]] $X$, contains $x$, and trivially intersects $A$. It is a [[open sets are unions of basis elements|union of basis elements]], none of which may intersect $A$ nontrivially but at least one of which, call it $B$, contains $x$. This suffices as the $B$ desired. Conversely, suppose there exists a [[neighborhood]] $U \ni x$ with $U \cap A= \emptyset$. $U$ is a [[open sets are unions of basis elements|union of basis elements]], at least one of which, call it $B$, contains $x$; note that $B \cap A = \emptyset$. Now, $X-B$ is [[closed set|closed in]] $X$ and contains $A$, but does not contain $x$ — as desired. > [!basicexample] > Let $X=\mathbb{R}$. If $A=(0,1]$, then $\overline{A}=[0,1]$, for any [[standard topology on the real line|basis element]] containing $0$ must intersect $A$ nontrivially (thus $0 \in \overline{A}$), while every point outside $[0,1]$ has a [[neighborhood]] disjoint from $A$. > > If $B=\left\{ \frac{1}{n} : n \in \mathbb{N} \right\}$, then $\overline{B}=\{ 0 \} \cup \{ B \}$. $0 \in \overline{B}$ because every [[neighborhood]] containing $0$ must nontrivially intersect an element of $B$ ([[archimedian property]]). Any element not in this set has a [[neighborhood]] disjoint of it: in the interesting case, the element is between two elements $\frac{1}{n}$ and $\frac{1}{n+1}$ of $B$... in which case taking the [[neighborhood]] $\left( \frac{1}{n+1}, \frac{1}{n+\frac{1}{2}} \right)$ suffices. > > If $C=\{ 0 \} \cup (1,2)$, then $\overline{C}=\{ 0 \} \cup [1,2]$ because every [[neighborhood]] containing $1$ or $2$ must nontrivially intersect $(1,2)$, while we can find a [[neighborhood]] around any element in the complement of $\{ 0 \} \cup [1,2]$ (an [[open set]]). > > If $D=\mathbb{Q}$, then $\overline{D}=\mathbb{R}$, since every [[open interval]] contains an element of $\mathbb{Q}$. > > If $E=\mathbb{R}_{+}$ is the set of positive reals, then $\overline{E}=\{ 0 \} \cup \mathbb{R}_{+}$. This is because every [[neighborhood]] of $\{ 0 \}$ intersects $\mathbb{R}_{+}$ nontrivially, while any point outside $\{ 0 \} \cup \mathbb{R}_{+}$ has a [[neighborhood]] disjoint from $\mathbb{R}_{+}$ because that complement is an [[open set]]. > > Consider the [[subspace topology|subspace]] $Y=(0,1]$ of $\mathbb{R}$. The set $A=\left( 0, \frac{1}{2} \right)$ is a subset of $Y$; its [[closure]] in $\mathbb{R}$ is $\left[ 0, \frac{1}{2} \right]$, its [[closure]] in $Y$ is $\left[ 0, \frac{1}{2} \right] \cap Y= (0, \frac{1}{2}]$. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```