----- > [!proposition] Proposition. ([[the sequential continuity lemma]]) > Let $X$ and $Y$ be [[topological space|topological spaces]] and $f:X \to Y$. If $f$ is [[continuous]], then for every [[converge|convergent]] [[sequence]] $(x_{n}) \to x$ in $X$, the image [[sequence]] $f(x_{n})$ converges to $f(x)$. The converse is not generally true, but it *is* true if $X$ is [[first-countable space|first-countable]] (e.g., [[metrizable]]). > [!proof]- Proof. ([[the sequential continuity lemma]]) > Let $f$ be [[continuous]] and $x_{n} \to x$. Let $V \subset Y$ be an open [[neighborhood]] containing $f(x)$. We want to find $N \in \mathbb{N}$ s.t. for all $n > N$ we have $f(x_{n}) \in V$. This is easy: since $f^{-1}(V)$ is an open neighborhood of $x$, there exists $N \in \mathbb{N}$ s.t. for all $n>N$ we have $x_{n} \in f^{-1}(V)$. It follows that $f(x_{n}) \in V$. Thus $f(x_{n}) \to f(x)$. Also see [[continuity preserves convergence]]. > > Conversely, suppose $X$ is [[metrizable]]; let the [[metric]] $d$ [[metric topology|induce]] the [[topological space|topology]] on $X$; and assume that the convergent sequence condition is satisfied. Let $A \subset X$; we show that $f(\overline{A}) \subset \overline{f(A)}$. If $x \in \overline{A}$, then by [[the sequence lemma]] (generalized to [[first-countable space|countability]] as is relevant) there exists a sequence $(x_{n})$ of points in $A$ converging to $x$. By assumption we have that $f(x_{n})$ converges to $f(x) \in f(A)$. [[the sequence lemma]] now tells us that $f(x) \in \overline{f(A)}$. We are done by [[continuity characterizations|closure characterization of continuity]]. ----- #### ---- #### 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 > ```