[[Topology of Metric Space]], [[Convergence, Completeness in Metric Space]] --- ### **Intro** We introduce some definitions and characterizes continuous mapping in abstract metric space. We do it from the ground up. When we talk about the distance ball, for a metric space $(X, d)$, we define: $ \begin{aligned} \mathbb B_r(\bar x|X) &= \{x\in X| d(x, \bar x) \le r\} \\ \overline{\mathbb B}_r(\bar x|X) &= \text{cl}(\mathbb B_r(\bar x|X)), \end{aligned} $ **Definition | Continuous Mapping Between Metric Space** > Let $(X, d_X), (Y, d_Y)$ be 2 metric spaces. The mapping $T: X\mapsto Y$ is continuous at $\bar x \in X$, if, for all $\epsilon > 0$ there exists $\delta > 0$ such that whenever $d_X(x, \bar x)\le \delta$, we have $d_Y(T\bar x, Tx) \le \epsilon$. Topologically, it can be expressed as: > $ > \forall \epsilon > 0 \exists \delta: T \mathbb B_\delta(\bar x|X) \subseteq \mathbb B_\epsilon(T\bar x|Y). > $ **Definition | Accumulation Point** > See [[Topology of Metric Space]] for the definition there. --- ### **Theorem | Open Sets Characterizations of Continuity** > Let $T$ be a mapping between the metric space $(X,d_X)$ and $(Y, d_Y)$, then it's a continuous mapping if and only if for all open set $O\in Y$, we have the pre-image $T^{-1}O$ is open in $X$. **Proof**: In $\implies$, we assume that the mapping $T$ is continuous and we show that if $A\subseteq Y$ is open set then $T^{-1}A$ is an open set as well. Let $\bar x \in T^{-1}A$, by the property of open we have: $ \begin{aligned} & \exists \epsilon >0: \mathbb B_{\epsilon}(T\bar x|Y) \subseteq A \quad \text{ By } A \text{ open} \\ & \exists \delta > 0: T \mathbb B_{\delta}(\bar x|X) \subseteq \mathbb B_\epsilon(T\bar x| Y) \subseteq A \quad \text{ By continuity of } T \\ \iff & \mathbb B_\delta(\bar x |X) \subseteq T^{-1}A, \end{aligned} $ which is the definition of $T^{-1}A$ is an open set. Otherwise for $\impliedby$ we consider a mapping whose pre-image always preserve openness of the set. To start, $\forall \epsilon > 0$, an open ball $\mathbb B_\epsilon(T\bar x|Y)$ is an open set. Let $y = T\bar x$ then we have: $ \begin{aligned} & y\in \mathbb B_{\epsilon}(T\bar x|Y) \\ \implies & x\in T^{-1}y \subseteq T^{-1}B_{\epsilon}(T\bar x|Y) \\ \implies & x\in T^{-1}B_{\epsilon}(T\bar x|Y) \\ & \exists \delta > 0: \mathbb B_{\delta}(\bar x| X) \subseteq T^{-1}B_{\epsilon}(T\bar x|Y), \end{aligned} $ where on the last line, we use the fact that $T^{-1}B_{\epsilon}(T\bar x|Y)$ is the pre-image of an open subset in $Y$, therefore, it's open. Combining everything together, we have recover the definition of a continuous mapping in metric space. **Remarks** A continuous function doesn't map an open set to an open set. Consider $x^2$ over the interval $(-1, 1)$, the range is $(1, 0]$. That is a closed and open set. --- ### **Compactness, continuity** --- ### **Examples 1 | Arc Length Linear Functional** We consider a linear functional applied to functions of $C^1[0, 1]$, and we show that an functional that returns the arc length, is not a continuous mapping. > The transformation $T: f \mapsto \int_{0}^1\sqrt{1 + (f')^2}dx$ (The arc length formula) in functional space $C^1[0, 1]$ equipped with the norm $\Vert \cdot\Vert_\infty$ is not continuous. One of the counter example one can use is this sequence of functions: > $ > \begin{aligned} > f_n(x) = \frac{\cos(2n\pi x)}{\sqrt{n}} > \end{aligned} > $ **Proof Priors** Before the proof, we establish the fact that $\sin(2n\pi x) \ge \frac{1}{\sqrt{2}}$ for all $x\in [\frac{1}{8n}, \frac{3}{8n}]$ and the fact that $f_n$ has a period of $1/n$. **Source**: This is exercise 2.3 in \<A friendly approach to functional analysis\>.