Definition
Topological Definition of Continuity
Consider two topological spaces and . A function is continuous if
\forall V \in \mathcal{T}_{Y},\ f^{-1}(V) \in \mathcal{T}_{X}\ &\Longleftrightarrow\ \forall B \in \mathcal{B}_{Y},\ f^{-1}(B) \in \mathcal{T}_{X}\\ &\Longleftrightarrow\ \forall A \subset X,\ f(\bar{A}) \subset \overline{f(A)}\\ &\Longleftrightarrow\ \forall (X\setminus B) \in \mathcal{T}_{Y},\ (X\setminus f^{-1}(B)) \in \mathcal{T}_{X}\\ &\Longleftrightarrow\ \forall x \in X,\ \exists \mathcal{N}_{x}\ \text{s.t.}\ f(\mathcal{N}_{x}) \subset \mathcal{N}_{f(x)} \end{aligned}$$ where $\mathcal{N}_{x}$ is a [[Neighborhood]] of $x$. A function is continuous if and only if the inverse image of any arbitrary open set in [[Codomain]] is an open set. ## Continuity of Real-Valued Function ### Continuous at a Point Suppose $f: \mathcal{D} \subset \mathbb{R}^{n} \to \mathbb{R}^{m}$, and $\mathbf{x}_{0} \in \mathcal{D}$ $$\begin{aligned} \lim\limits_{\mathbf{x} \to \mathbf{x}_{0}} f(\mathbf{x}) = f(\mathbf{x}_{0}) &\Leftrightarrow \forall \epsilon > 0, \exists \delta > 0, \forall x \in \mathcal{D}, (|\mathbf{x}-\mathbf{x}_{0}| < \delta \Rightarrow |f(\mathbf{x})-f(\mathbf{x}_{0})| < \epsilon)\\ &\Leftrightarrow \forall\epsilon>0, \exists\delta>0\ \text{s.t}\ f(B_{\delta}(\mathbf{x})) \subset B_{\epsilon}(f(\mathbf{x}))\\ &\Leftrightarrow \forall\epsilon>0, \exists\delta>0\ \text{s.t}\ B_{\delta}(\mathbf{x}) \subset f^{-1}(B_{\epsilon}(f(\mathbf{x}))) \end{aligned}$$ $f$ is continuous at a point $\mathbf{x}_{0}$ if the [[Limit of a Function|limit]] of $f(\mathbf{x})$, as $\mathbf{x}$ approaches $\mathbf{x}_{0}$, exists and is equal to $f(\mathbf{x}_{0})$ ### Continuous on an Open Interval Suppose $f: \mathcal{D} \subset \mathbb{R}^{n} \to \mathbb{R}^{m}$, and $X \subset \mathcal{D}$ $$\forall \mathbf{x}_{0} \in X, \lim\limits_{\mathbf{x} \to \mathbf{x}_{0}} f(\mathbf{x}) = f(\mathbf{x}_{0})$$ A function is continuous at every point in an open interval $X$ ### Continuous Function Suppose $f: \mathcal{D} \subset \mathbb{R}^{n} \to \mathbb{R}^{m}$ $$\forall \mathbf{x}_{0} \in \mathcal{D},\ \lim\limits_{\mathbf{x} \to \mathbf{x}_{0}} f(\mathbf{x}) = f(\mathbf{x}_{0})$$ A function is continuous at every point in its domain ### Right-Continuous Suppose $f: \mathcal{D} \to \mathbb{R}$, and $a \in \mathcal{D}$ $$\forall \epsilon > 0, \exists \delta > 0, \forall x \in \mathcal{D}, (0 \leq x-a < \delta \Rightarrow |f(x)-f(a)| < \epsilon) \Leftrightarrow \lim\limits_{x \to a^{+}} f(x) = f(a)$$ $f$ is right-continuous at $a$ The [[Limit of a Function#right-sided-limit|right-sided limit]] of $f(x)$, as $x$ approaches $a$ from the right side, exists and is equal to $f(a)$ ### Left-Continuous Suppose $f: \mathcal{D} \to \mathbb{R}$, and $a \in \mathcal{D}$ $$\forall \epsilon > 0, \exists \delta > 0, \forall x \in \mathcal{D}, (0 \leq a-x < \delta \Rightarrow |f(x)-f(a)| < \epsilon) \Leftrightarrow \lim\limits_{x \to a^{-}} f(x) = f(a)$$ $f$ is left-continuous at $a$ The [[Limit of a Function#left-sided-limit|left-sided limit]] of $f(x)$, as $x$ approaches $a$ from the left side, exists and is equal to $f(a)$ # Properties ## Construction of Continuous Functions Suppose $a \in \mathcal{D}$, and $f, g: \mathcal{D} \to \mathbb{R}$ is continuous at $a$ Then, the following are satisfied - $f + g$ is continuous at $a$ - $f - g$ is continuous at $a$ - $f \cdot g$ is continuous at $a$ - $g(a) \neq 0 \Rightarrow \cfrac{f}{g}$ is continuous at $a$ # Facts ![[Lipschitz Continuity#^e0a56a|^e0a56a]] > Every continuous function $f: [a, b] \to \mathbb{R}$ is [[Darboux Integral|integrable]] ^c92817 ![[Heine-Cantor Theorem]] > Given two continuous functions $f: D_{f} \subset \mathbb{R}^{n} \to R_{f} \subset D_{g}$ and $g: D_{g} \subset \mathbb{R}^{m} \to R_{g} \subset \mathbb{R}^{l}$, then their composition $h := g \circ f: D_{f} \subset \mathbb{R}^{n} \to R_{g} \subset \mathbb{R}^{l}$ is continuous. > A [[Constant Function]] $f: X \to Y,\quad f(x)=y_{0}$ is continuous. > [[Inclusion Function|Inclusion Map]] $j: A \hookrightarrow X,\quad j(x) = x$ where $A \subset X$, is continuous > $(f: X \to Y)\in C^{0}, (g: Y \to Z)\in C^{0}\ \Rightarrow\ (g\circ f: X \to Z) \in C^{0}$ > A [[Function Composition|composite function]] of continuous functions is continuous > $(f: X \to Y) \in C^{0}, A \subset X\ \Rightarrow\ (f|_{A}: A \to Y) \in C^{0}$ > A continuous function with restricted [[Domain]] is continuous > $(f: X \to Y) \in C^{0}, f(x) \subset Z \subset Y\ \Rightarrow\ (g: X \to Z,\quad g(x)=f(x)) \in C^{0}$ > $(f: X \to Y) \in C^{0}, Y \subset Z\ \Rightarrow\ (h: X \to Z,\quad h(x)=f(x)) \in C^{0}$ > A continuous function with expanded and restricted [[Codomain]] is continuous. > $(f|_{U_\alpha}: U_{\alpha}\to Y\ \text{s.t.}\ \bigcup_{\alpha} U_{\alpha} = X) \in C^{0} \Rightarrow (f: X \to Y) \in C^{0}$ > Consider a collection $(U_\alpha)$ of [[Open Set|open sets]] in $X$. If $X = \bigcup_{\alpha} U_{\alpha}= X$ and $f|_{U_\alpha}$ is continuous, then $f: X \to Y$ is continuous. ![[Gluing Lemma]] > Consider [[Topological Space|topological Spaces]] $(A, \mathcal{T}), (X \times Y, \mathcal{T})$ and a function $f: A \to X\times Y,\quad f(a) = (f_{1}(a), f_{2}(a))$. > $f \in C^{0} \Leftrightarrow (f_{1}: A \to X), (f_{2}: A \to Y) \in C^{0}$ > A function with a [[Product Topology]] [[Codomain]] is continuous if and only if all of its coordinate functions are continuous. > Consider [[Topological Space|topological Spaces]] $(A, \mathcal{T}), (\prod\limits_{i \in \mathbb{N}}X_{i}, \mathcal{T})$ and a function $f: A \to \prod\limits_{i \in \mathbb{N}}X_{i},\quad f(a) = (f_{1}(a), f_{2}(a), \dots, f_{n}(a), \dots)$. > $f \in C^{0} \Leftrightarrow \forall i \in \mathbb{N},\ (f_{i}: A \to X_{i}), \in C^{0}$ > A function with a countably infinite [[Product Topology]] [[Codomain]] is continuous if and only if all of its coordinate functions are continuous. ![[Closed Map#^b9de02|^b9de02]]