> [!definitionb] Definition > > Let $(X, \topo_X)$ and $(Y, \topo_Y)$ be [[Topological Space|topological spaces]] and $f: X \to Y$ be a [[Function|function]]. Then $f$ is **continuous** if the preimage $f^{-1}(U)$ of any [[Open Set|open set]] in $Y$ is also open in $X$. > [!definition] > > Let $(X, \topo_X)$ be a [[Topological Space|topological space]] and $f: X \to Y$, then $f$ is **continuous at** some $x \in X$ if the preimage of any [[Neighbourhood|neighbourhood]] $V$ of $f(x)$ is a neighbourhood of $x$. > [!theorem] > > Let $(X, \topo_X)$ and $(Y, \topo_Y)$ be topological spaces and $f: X \to Y$, then $f$ is continuous if and only if $f$ is continuous at every $x \in X$. > > *Proof*. Suppose that $f$ is continuous. Let $x \in X$ and $V$ be a neighbourhood of $f(x)$, then there exists an open set $U$ such that $f(x) \in U \subset V$. This means that $f^{-1}(V) \supset f^{-1}(U) \ni x$ and as $f^{-1}(U)$ is an open set, $f^{-1}(V)$ is a neighbourhood of $x$. > > Suppose that $f$ is continuous at every $x \in X$. Let $U \in \topo_Y$ be an open set, then $U$ is a neighbourhood for all $f(x) \in U$. As a result, $f^{-1}(U)$ is a neighbourhood for every $x \in f^{-1}(U)$. This means that $U$ is its own interior and therefore is open. > [!theorem] > > Let $X$ and $Y$ be topological spaces and $f: X \to Y$. Then $f$ is continuous at $x \in X$ if and only if for every net $\net{x}$ converging to $x$, $\angles{f(x_\alpha)}$ converges to $f(x)$. > > *Proof*. Suppose that $f$ is continuous at $x$, then for any $V \in \cn(f(x))$, $f^{-1}(V) \in \cn(x)$. If $x_\alpha \to x$, then $\angles{x_\alpha}$ is eventually in $f^{-1}(V)$, and $\angles{f(x_\alpha)}$ is eventually in $V$. Therefore $f(x_\alpha) \to f(x)$. > > Suppose that $f$ is not continuous at $x$, then there exists $V \in \cn(f(x))$ such that $f^{-1}(V) \not\in \cn(x)$, meaning that $x \not\in (f^{-1}(V))^o$ and $x \in \ol{f^{-1}(V)^c} = \ol{f^{-1}(V^c)}$. This allows the construction of a net $\angles{x_\alpha}$ in $f^{-1}(V^c)$ that converges to $x$. However, since $f(x_{\alpha}) \not\in V$ for all $\alpha$, $f(x_\alpha) \not\to f(x)$. > [!theorem] > > Let $(X, \topo_X)$ be a topological space, $Y$ be a set, $\ce \subset \pow{Y}$ be a family of subsets and $\topo_Y = \topo(\ce)$ be the topology generated by $\ce$, then $f: X \to Y$ is continuous if and only if $f^{-1}(V)$ is open in $X$ for each $V \in \ce$. > > *Proof*. If $f$ is continuous, then $\ce \subset \ct_Y$ implies that $f^{-1}(V) \in \topo_X$ for every $V \in \ce$. > > Now suppose that $f^{-1}(V)$ is open in $X$ for each $V \in \ce$. Let > $ > \cf = \bracs{\bigcap_{i = 1}^{n}V_i: V_i \in \ce, n \in \nat} > $ > be the family of finite intersections of $\ce$ sets, then > $ > f^{-1}\paren{\bigcap_{i = 1}^{n}V_i} = \bigcap_{i = 1}^{n}f^{-1}(V_i) \in \topo_X > $ > their preimages are also open. Since > $ > \topo_Y = \bracs{\bigcup_{U \in \mathcal{U}}U: \mathcal{U} \subset \cf} > $ > is the collection of arbitrary unions of $\cf$ sets, > $ > f^{-1}\paren{\bigcup_{U \in \mathcal{U}}U} = \bigcup_{U \in \mathcal{U}}f^{-1}(U) \in \topo_X > $ > their preimages are also open. Therefore $f$ is continuous. > [!definition] > > A [[Function|function]] $f: D \to M^\prime$, $D \subseteq M$ between [[Metric Space|metric spaces]] is continuous over a set $S \subseteq D$ if the [[Limit|limit]] of the function is equal to the function of the limit > $ > \forall c \in S, \lim_{x \to c}f(x) = f(c) > $ > More formally, > $ > \forall c \in S, \varepsilon > 0,\exists \delta > 0: > |x - c| < \delta \Rightarrow > |f(x) - f(c)| < \varepsilon > $ > This requires $c$ to be a [[Cluster Point|cluster point]]. If $c$ is not a cluster point (isolated) then the function is automatically continuous there. > [!theorem] > > The following definitions of continuity are equivalent: > - $f$ is continuous ([[Topological Space|Topology]]) > - $\lim_{x \to c}f(c) = f(c)$ ([[Limit]]) > - For any [[Sequence|sequence]] $(x_n)$ in $D$: $\limv{n}x_n = c$ implies $\lim_{n \to \infty}f(x_n) = f(c)$. (Sequential) > > *Proof*. Suppose that $\lim_{x \to c}f(c) = f(c) \forall c \in D$, and that $f(D)$ is open, meaning > $ > \forall \varepsilon > 0, c \in D, \exists \delta > 0: |x - c| < \delta \Rightarrow |f(x) - f(c)| < \varepsilon > $ > and > $ > \forall f(x) \in D, \exists \varepsilon > 0: \bracs{y: |y - f(x)| < \varepsilon} \subseteq M^\prime > $ > then let $x \in D$, the preimage of the set $\bracs{y: |y - f(x)| < \varepsilon} \subseteq M^\prime$ would contain $\bracs{z: |z - x| < \delta}$. Therefore $D$ would also be an open set. > > Suppose that the preimage of every open set is also an open set, then obviously > $ > \forall y \in \bracs{y: |y - f(x)| < \varepsilon}, \exists \delta > 0: \bracs{z: |z - x| < \delta} \subseteq f^{-1}(D) > $ > > The limit definition easily implies the sequential definition. > > Suppose that the sequential definition is true but the limit definition is false, then > $ > \exists \varepsilon > 0: \forall \delta > 0, |x - c| < \delta \Rightarrow |f(x) - f(c)| \ge \varepsilon > $ > However, any sequence would contradict this. Therefore the limit definition holds. > [!theorem] > > Let $(X, \topo_X)$, $(Y, \topo_Y)$ be topological spaces. A function $f: X \to Y$ is continuous if and only if there exists $U, V \in \topo_X$ such that $X = U \cup V$ and $f$ is continuous in the [[Relative Topology|relative topology]] on both of them. > > *Proof*. Suppose that $f$ is continuous in the induced topology of $U$ and $V$, then for any open $E \in \topo_Y$, $f^{-1}(E)$ is open in $U$ and $V$. > > Since $U \cup V = X$ and $f^{-1}(E) \cap U$ is equal to $W \cap U$ for some $W \in \topo_X$, we can write > $ > f^{-1}(E) = [f^{-1}(E) \cap U] \cup [f^{-1}(E) \cap V] > $ > as a union of open sets. Therefore $f$ is continuous.