Properties:: [[constant function is continuous]], [[inclusion map is continuous]], [[composition of continuous functions is continuous]], [[domain restriction preserves continuity]], [[codomain restriction and expansion of continuous function is continuous]], [[local formulation of continuity]] Sufficiencies:: *[[Sufficiencies]]* Equivalences:: [[continuous#^51cda5|Basis equivalence]], [[continuous#^356548|subbasis equivalence]], [[metric space characterization of continuity]], [[continuity characterizations|closed set characterization of continuity]], [[continuity characterizations|closure characterization of continuity]], [[continuity characterizations|nestling characterzation of continuity]] ---- - Let $X,Y$ be [[topological space|topological spaces]]. > [!definition] Definition. ([[continuous]]) > A map $f: X \to Y$ is called **continuous** if [[inverse image|inverse images]] of [[open set|open sets]] in $Y$ are [[open set|open sets]] in $X$. > The [[category]] $\mathsf{Top}$ has [[topological space|topological spaces]] as objects and continuous maps as morphisms. When working over $\mathsf{Top}$, often one just says 'map' instead of 'continuous map'. > [!justification] > This definition captures, among other things, the essence of function [[continuous|continuity]] in [[metric space|metric spaces]], without actually needing a [[metric]]. ^justification > [!equivalence] [[basis for a topology|Basis]] Equivalence. > If the [[topological space|topology]] of the range space $Y$ is [[topology generated by a basis|given by a basis]] $\mathscr{B}$, then $f:X \to Y$ is [[continuous]] *if and only if* the [[inverse image]] of every [[basis for a topology|basis element]] of $Y$ is [[open set|open in]] $X$. > \ > **Proof.** One direction is immediate ,since [[basis for a topology|basis elements are open]]. So, assume the [[inverse image]] of every [[basis for a topology|basis element of]] $Y$ is [[open set|open in]] $X$. Then consider an arbitrary [[open set]] in $Y$, written as a [[open sets are unions of basis elements|union of basis elements]] $\bigcup_{\alpha}^{}B_{\alpha}$. By [[preimages and unions commute]] we have $f^{-1}(\bigcup_{\alpha}{B_{\alpha}}) = \bigcup_{\alpha}^{} f ^{-1}(B_{\alpha}),$ > and the RHS is open in $X$ since each $B_{\alpha}$ is. ^51cda5 > [!equivalence] [[subbasis for a topology|Subbasis]] Equivalence. > If the [[topological space|topology]] of the range space $Y$ is [[subbasis for a topology|subbasis]] $\mathscr{S}$, then $f:X \to Y$ is [[continuous]] *if and only if* the [[inverse image]] of every [[subbasis for a topology|subbasis element]] of $Y$ is [[open set|open in]] $X$. > \ > **Proof.** One direction is immediate, since [[subbasis for a topology|subbasis elements are open]]. So, assume the [[inverse image]] of every [[subbasis for a topology|subbasis element]] is [[open set|open in]] $Y$. An element of $Y$ can be written as a union of finite intersections of elements of $\mathscr{S}$. Then, recall that ([[subbasis for a topology#^c11575|via the subbasis justification proof]]) the set obtained by taking finite intersections of elements in $\mathscr{S}$ is a [[basis for a topology|basis generating]] $Y$. The plan is thus to show that the [[inverse image]] of each intersection $S_{1} \cap \dots \cap S_{n}$ is [[open set|open in]] $X$. [[preimages and intersections commute|Because]] $f^{-1}(S_{1} \cap \dots \cap S_{n})=f^{-1}(S_{1}) \cap \dots \cap f^{-1}(S_{n})$ > we're done. ^356548 > [!warning] The [[topological space|topologies]] matter! > Let $\mathbb{R}$ denote the [[real numbers|real line]] with the [[standard topology on the real line|standard topology]] and $\mathbb{R} _\ell$ the [[real numbers|real line]] with the [[lower limit topology, upper limit topology|lower-limit topology]]. Then the identity function $f: \mathbb{R} \to \mathbb{R}_\ell$ is *not* [[continuous]] on $\mathbb{R}$, because the [[inverse image]] of the [[open set]] $[a,b)$ of $\mathbb{R} _\ell$ is not [[open set|open in]] $\mathbb{R}$. > Consider the [[finite complement topology|cofinite topology]] on the real line. Take $f(x)=\sin x: \mathbb{R}^{\text{cofinite}} \to \mathbb{R}^{\text{cofinite}}$. $f$ is not [[continuous]]; indeed, for the open subset $U:=\mathbb{R}-\{ 0 \}$, we have that $\mathbb{R}-U=\{ 0 \}$ is finite, however $f^{-1}(U)=\{ x \in \mathbb{R}: \sin x \neq 0 \}$ and thus $\mathbb{R}-f^{-1}(U)= \{ x \in \mathbb{R} : \sin x =0 \}=\{ n\pi : n \in \mathbb{Z} \}$ which is not finite. ^nonexample > [!basicexample] > Let $X:=\{ 0,1 \}$ and consider the topology $\tau=\{ \emptyset, \{ 0 \}, X \}$. Then any [[continuous]] function $f:X \to \mathbb{R}$ ($\mathbb{R}$ with the standard topology) is a constant, i.e., $f(0)=f(1)$ \ Indeed, contrapositively suppose $f$ is not constant: $f(0) \neq f(1)$. Letting $\varepsilon:= |f(0)-f(1)|$, obtain an [[open ball]] $B:=B_{\varepsilon}\big( f(1) \big) \subset \mathbb{R}$. We can conclude the following: > >- $f^{-1}\big( B \big) \neq \emptyset$, since $1 \in f^{-1}(B)$; >- $0 \notin f^{-1}(B)$, since $B \cap \{ f(0) \} = \emptyset$, implying $f^{-1}(B) \neq \{ 0 \}$ and $f^{-1}(B) \neq X$. > Thus $f^{-1}(B) \notin \tau$ and we so $f$ is not [[continuous]] wrt the given [[topological space|topologies]]. ^f23fad > [!specialization] > [[metric-space continuity]] ^specialization ---- #### ---- #### 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 > ``` --- # Legacy Notes # Math 395 Munkres General Definition Let $X$ and $Y$ be [[metric space]]s with [[metric]]s $d_{X}$ and $d_{Y}$ respectively. We say that a function $f: X \to Y$ is *continuous at a point* $x_{0} \in X$ if for each [[open set]] $V$ of $Y$ containing $f(x_{0})$ there exist an [[open set]] $U$ of $X$ s.t. $f[U] \subset V$. $f$ is called continuous if it is continuous at every point. #### Equivalently $f$ is [[continuous]] iff for each [[open set]] $V$ of $Y$, the set $f^{-1}(V) = \{x \in X \vert f(x)\in V \}$ is *open in *. Alternatively, $f$ is [[continuous]] iff for each [[closed set]] $D$ of $Y$, the set $f^{-1}(D) = \{x \in X \vert f(x)\in D \}$ is *closed in *. # Math 395 Munkres [[Metric]]s-Specific Formulation The function $f$ is [[continuous]] at $x_{0}$ iff the following holds— *for each $\varepsilon>0$, there exists $\delta>0$ such that* $d_{Y}(f(x),f(x_{0})) < \varepsilon \text{ whenever } d_{X}(x, x_{0}) < \delta$ ## Important Lemma 1. *If* $p$ is an [[isolated point]] of domain $X$ , then $f$ is [[continuous]] at $p$ 2. *Else*, $f$ is continuous at $p$ **iff** $\lim_{ x \to p } f(x)=f(p)$ # Operations on Continuous functions ## Composition Let $f: X \to Y$ and $g: Y \to Z$. If $f$ is continuous at $x_{0}$ and $g$ is continuous at $y_{0}=f(x_{0})$, then $g \circ f$ is continuous at $x_{0}$. #### Proof Since $g$ is continuous at $x_{0}$ we have that $\forall \varepsilon>0$ $\exists \delta_{Y}>0$ s.t. (1) $d_{Z}(g(f(x)),g(f(x_{0}))) < \varepsilon \text{ whenever } d_{Y}(f(x), f(x_{0})) < \delta_{Y}.$ Likewise, since $f$ is continuous at $x_{0}$ we have that $\forall \varepsilon_{Y}>0$ $\exists \delta>0$ s.t. (2) $d_{Y}(f(x),f(x_{0})) < \varepsilon_{Y} \text{ whenever } d_{X}(x, x_{0}) < \delta.$ So fix $\varepsilon>0$. Obtain the corresponding $\delta_{Y}$ s.t. (1) holds. Since (2) holds no matter $\varepsilon_{Y}$, it holds for $\varepsilon_{Y} = \delta_{Y}$. Obtain the corresponding $\delta$ s.t. (2) holds and we have what we want— that $\forall \varepsilon>0 \ \exists \delta>0$ s.t. $d_{Z}(g(f(x)),g(f(x_{0}))) < \varepsilon \text{ whenever } d_{X}(x, x_{0}) < \delta.$ ## Componentwise Continuity Let $X$ be a [[metric space]]. Let $f : X \to \mathbb{R}^n$ be have the form $f(x) = (f_{1}(x), \dots, f_{n}(x))$ Then $f$ is [[continuous]] at $x_{0}$ **iff** each function $f_{i}: X \to \mathbb{R}$ is continuous at $x_{0}$. The functions $f_{i}$ are called the *component functions* of $f$. ## Algebraic Manipulations Let $f,g$ be continuous at $x_{0}$. Then $f+g$, $f-g$, and $fg$ are continuous at $x_{0}$ as well, as is $\frac{f}{g}$ provided $g(x_{0})\neq 0$. ## Cool Part Putting these **composition** and **algebraic** properties together allows us to immediately conclude that a function such as $f(s,t,u,v) = \frac{\sin(s+t)}{e^{uv}}$ from is continuous in $\mathbb{R}^{4}$. In general, **functions formed from the familiar real-valued continuous functions of calculus, using algebraic operations and composites, are continuous in $\mathbb{R}^n$ **. # More Equivalencies - half the battle of proving something [[continuous]] is choosing the best equivalent theorem (or definition) to show! ### With Metrics - We can equivalently **define** continuity for a function between [[metric space]]s like so: - Let $X,Y$ be [[metric space]]s, let $p\in X$. $f: X \to Y$ is continuous at $p$ **if(f)** for all $\varepsilon>0$ there exists $\delta>0$ s.t. $x \in U_{x}(p;\delta) \implies f(x) \in U_{Y}(f(p);\varepsilon).$ - - If $p \in X$ is not an [[isolated point]], then $f: X \to Y$ is continuous at $p$ **iff** $\lim_{ x \to p }f(x) = f(p). $ Moreover, the above in turn holds **iff** $\lim_{ x \to p } d_{Y}(f(x),f(p)) = 0,$ that is, **iff** $\lim_{ d_{X}(x,p) \to 0 } d_{Y}(f(x),f(p)) = 0.$ # with [[sequence]]s ([[math 297]]) Suppose $(V, \langle , \rangle_V)$ and $(W, \langle , \rangle_W)$ are [[inner product space]]s (more generally, [[normed vector space]]s should be fine). Suppose $A \subset V$, $\vec a \in A$, and $\vec f: A \to W$ is a function. $\vec f$ is [[continuous]] at $\vec a$ **iff** for all [[sequence]]s $(\vec a_n)$ in $A$ that [[converge]] to $\vec a$ we have that the [[sequence]] $\big(\vec f( \vec a_n) \big )$ [[converge]]s to $\vec f(\vec a)$. ^definition-1