----- > [!proposition] Proposition. ([[the sequence lemma]]) > Let $X$ be a [[topological space]]; let $A \subset X$. If there is a [[sequence]] $(x_{n})_{n \in \mathbb{N}}$ of points of $A$ [[converge|converging]] to $x$, then $x \in \overline{A}$. The converse holds if $X$ is [[first-countable space|first countable]] (e.g., [[metrizable]]). ^181237 > [!proposition] Corollary. > Call a subset $A \subset X$ **sequentially closed** if every [[sequence]] in $A$ [[converge|converges]] to an element of $A$. The sequence lemma implies that > > $(\text{closed }\implies \text{sequentially closed}) \text{ always}$ > and > > $(\text{closed} \iff \text{sequentially closed}) \text{ if }X \text{ is first-countable}.$ > > > [!proof]- Proof. ([[the sequence lemma]]) > ~ > > First, for [[metrizable]] spaces: > Let $(x_{n})_{n \in \mathbb{N}}$ be a [[sequence]] of points in $A$ converging to $x$. Suppose $x \notin \overline{A}$. Then there exists an open [[neighborhood]] $B \subset X$ satisfying $x \in B \subset X \cut A$. Since $(x_{n}) \to x$, we may fix $N \in \mathbb{N}$ s.t. for all $n > N$ we have $x_{n} \in B$. But we already stated $x_{n} \in A$ and $B \cap A = \emptyset$. Contradiction. > > Conversely suppose $X$ is [[metrizable]]; suppose it is [[metric topology|induced]] by [[metric]] $d$. Fix $x \in \overline{A}$. Then we can construct a sequence $(x_{n})$ converging to $x$ by choosing $x_{n}$ out of $B_{\frac{1}{n}}(x) \cap A$ at each step (we know we can do this because of the [[neighborhood-basis characterization of set closure]]). Indeed, for any open [[neighborhood]] $U$ containing $x$ we can find a basis element $B_{\varepsilon}(x) \subset U$ for some $\varepsilon>0$; then the [[archimedian property]] tells us that there exists $N$ large enough that $B_{\frac{1}{n}}(x) \subset B_{\varepsilon}(x)$ for all $n < N$— implying $x_{n} \in B_{\frac{1}{n}}(x) \subset B_{\varepsilon}(x) \subset U$ > and since $U$ was arbitrary we conclude $x_{n} \to x$. > > Conversely, suppose $X$ is [[first-countable space|first-countable]]. Fix $x \in \overline{A}$. Using the **lemma**, Obtain a [[sequence]] of open neighborhoods $(U_{n})_{n \in \mathbb{N}}$ of $x$ such that for any neighborhood $N$ of $x$ there exists $n_{0}$ s.t. $U_{n} \subset N$ for any $n \geq n_{0}$. Then we can construct a sequence $(x_{n})$ converging to $x$ by choosing $x_{n}$ out of $U_{n} \cap A$ (we can just 'choose' $x_{n}$ out of $U_{n} \cap A$ by using the [[neighborhood-basis characterization of set closure]]). To see convergence, observe that for any open [[neighborhood]] $N$ containing $x$ we can find $n_{0} \in \mathbb{N}$ such that for all $n \geq n_{0}$ we have $x_{n} \subset U_{n} \subset N.$ > and since $N$ was arbitrary we conclude $x_{n} \to x$. > > **Lemma.** (might be overkill, idk it was just part a) > Let $X,Y$ be [[topological space|topological spaces]] such that $X$ is [[first-countable space|first-countable]]. > > **a. Prove that there exists a [[sequence]] of open neighborhoods $(U_{n})_{n}$ of $x$ s.t. for any neighborhood $N$ of $x$ there exists $n_{0}$ s.t. $U_{n} \subset N$ for any $n \geq n_{0}$.** > > $X$ is first-countable; using this, fix a *countable* collection $\mathscr{B}$ of open neighborhoods of $x$ s.t. each neighborhood of $x$ contains at least one of the elements of $\mathscr{B}$. $\mathscr{B}$ is countable; enumerate it as the [[sequence]] $(B_{i})_{i \in \mathbb{N}}$. Then we define $(U_{n})_{n}$ by setting $U_{n}:= B_{1} \cap \dots \cap B_{n}.$(note $U_{n}$ is an open set as a finite intersection thereof). Observe that $U_{1} \supset U_{2} \supset \dots$. Now fix an open [[neighborhood]] $N \ni x$. Using first-countability there exists some $B_{n_{0}}$, $n_{0} \in \mathbb{N}$, s.t. $x \in B_{n_{0}} \subset N$. Now $U_{n_{0}}=B_{1} \cap \dots \cap B_{n_{0}} \subset B_{n_{0} } \subset N$ > and we are done. ----- #### ---- #### 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 > ```