----- Let $A$ be a subset of a [[topological space]] $X$. > [!proposition] Proposition. ([[closure is set together with limit points]]) > Let $A'$ be the set of all [[limit point]]s of $A$. Then $\overline{A}=A \cup A'.$ > \ > **Corollary.** $A$ is [[closed set|closed in]] $X$ if and only if it contains all its [[limit point]]s. > [!proof]- Proof. ([[closure is set together with limit points]]) > We'll show two-way inclusion. If $x \in A'$, by definition every [[neighborhood]] of $x$ intersects $A$ nontrivially (at a point other than $x$ itself). Hence $x \in \overline{A}$; so, $A' \subset \overline{A}$. And we know that $A \subset \overline{A}$. Hence $A \cup A' \subset \overline{A}$. \ Now for the reverse inclusion: suppose $x \in \overline{A}$. Then every [[neighborhood]] of $x$ intersects $A$ nontrivially. If $x\in A$, this intersection clearly contains points other than $x$ and we have $x \in A'$ (and thus $x \in A \cup A'$). Else $x \in \overline{A} - A$. In this case we have $x \notin A$ and that every [[neighborhood]] of $x$ intersects $A$ nontrivially— meaning that every [[neighborhood]] of $x$ intersects $A$ at a point other than $x$ itself. Hence $x \in A'$ and thus in $A \cup A'$. This means $\overline{A} \subset A \cup A'$. > [!proof]- Proof of Corollary. (closed iff contains all limit points) > $A$ is [[closed set|closed in]] $X$ if and only if $A=\overline{A}$, while by the **above** $A=\overline{A}$ if and only if $A \subset A'$. > [!proof] Proof in Euclidean Space of Corollary. > Here $X$ is a subset of euclidean space and $A \subset X$. $\to.$ Suppose that $A$ is [[closed set|closed in]] $X$, i.e., $X-A$ is [[topological space|open in]] $X$. Let $(x_{n})_{n \in \mathbb{N}}$ be a [[converge|convergent]] [[sequence]] of points in $A$ with limit $x$; we want to show $x \in A$. Suppose $x \notin A$. Then $x \in X-A$ and because $X-A$ is open we can find a ball $B_{\varepsilon}(x)\subset X-A$ of radius $\varepsilon > 0$ containing it. Since $(x_{n})$ converges, there exists $N > {N}$ such that for all $n > \mathbb{N}$ we have $x_{n} \in B_{\varepsilon}(x) \subset X-A$. For such $n$ we simultaneously have $n \in A$ and $n \in X-A$, a contradiction. Thus $x \in A$. > $\leftarrow$. Conversely, suppose every convergent sequence in $A$ converges to an element of $A$. We want to show $X-A$ is open. Suppose it is not. Then there exists $x \in X-A$ s.t. for all $\varepsilon>0$ we have $a \in B_{\varepsilon}(x)$ for some $a \in A$. In particular, we may construct a sequence $(a_{n})_{n \in \mathbb{N}}$ such that $a_{n} \in A \cap B_{\frac{1}{n}}(x)$ by taking $\varepsilon = \frac{1}{n}$ for each $n \in \mathbb{N}$. This is a sequence of elements in $A$ converging to $x \notin A$, a contradiction. ----- #### ---- #### 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 > ```