% vim: tw=50 % 20/10/2022 11AM \subsubsection*{Remarks} \begin{enumerate}[(1)] \item Don't have a nice rephrasing of (i) in terms of similar concepts in the reals. Would want to write ``$e(f(x), b) \to 0$ as $d(x, a) \to 0$'', but this is meaningless. \item (i) says nothing about what happens at the point $a$ itself. For example, let $f : \RR \to \RR$ \[ f(x) = \begin{cases} 1 & x = 0 \\ 0 & x \neq 0 \end{cases} \] Then $f(x) \to 0$ as $x \to 0$ (but $f(0) \neq 0$ so $f$ is not continuous at 0). If we have that $f$ is continuous then \[ d(x, a) = 0 \implies x = a \implies f(x) = f(a) = e(f(x), f(a)) = 0 \] so we can drop `$0 <$' from the definition when we come to define continuity. \item Can rewrite definition (v): $f$ is continuous on $W$ if and only if $f \mid_W$ is a continuous function $f \mid_W : W \to Y$ thinking of $W$ as a subspace of $X$. That is \[ \forall a \in W ~ \forall \eps > 0 ~ \exists \delta > 0 ~ \forall x \in W \quad d(x, a) < \delta \implies e(f(x), f(a)) < \eps \] In particular, note the subtlety that this \emph{only} mentions points of $W$. So, under this definition, for example $f : \RR \to \RR$, \[ f(x) = \begin{cases} 1 & x \in [0, 1] \\ 0 & x \not\in [0, 1] \end{cases} \] then $f$ is continuous on $[0, 1]$, but $f$ is not continuous at points 0 and 1. \end{enumerate} \begin{flashcard} \begin{proposition} Let $(X, d)$, $(Y, e)$ be metric spaces. Let $f : X \to Y$ and $a \in X$. Then $f$ is continuous at $a$ if and only if \cloze{whenever $(x_n)$ is a sequence in $X$ with $x_n \to a$ then $f(x_n) \to f(a)$.} \end{proposition} \end{flashcard} \begin{proof} \begin{enumerate}[(1)] \item[$\rightarrow$] Suppose $f$ is continuous at $a$. Let $(x_n)$ be a sequence in $X$ with $x_n \to a$. Let $\eps > 0$. As $f$ continuous at $a$ we can find $\delta > 0$ such that $\forall x \in X$, $d(x, a) < \delta \implies e(f(x), f(a)) < \eps$. As $x_n \to x$ we can find $N$ such that $n \ge N \implies d(x_n, a) < \delta$. Let $n \ge N$. Then $d(x_n, a) < \delta$ so $e(f(x_n), f(a)) < \eps$. Hence $f(x_n) \to f(a)$. \item[$\Leftarrow$] Suppose $f$ is not continuous at $a$. Then there is some $\eps > 0$ such that $\forall \delta . 0$, $\exists x \in X$ with $d(x, a) < \delta$ but $e(f(x), f(a)) \ge \eps$. Now take $\delta = \frac{1}{1}, \half, \frac{1}{3}, \dots$ we obtain a sequence $(x_n)$ with, for each $n$ \[ d(x_n, a) < \frac{1}{n} \quad \text{and} \quad e(f(x_n), f(a)) \ge \eps .\] Hence $x_n \to a$ but $f(x_n) \not\to f(a)$. \end{enumerate} \end{proof} \begin{proposition} Let $(W, c), (X, d), (Y, e)$ be metric spaces, let $f : W \to X$, let $g : X \to Y$ and let $a \in W$. Suppose $f$ is continuous at $a$ and $g$ is continuous at $f(a)$. Then $g \circ f$ is continuous at $a$. \end{proposition} \begin{hiddenflashcard}[composition-of-continuous-is-continuous] Prove that composition of continuous functions (in a metric space) is continuous. \\ \cloze{ If you are allowed to quote the fact a function is continuous if and only if $f(x_n) \to f(x)$ whenever $x_n \to x$, then this gives a very quick proof of this fact. } \end{hiddenflashcard} \begin{proof} Let $(x_n)$ be a sequence in $W$ with $x_n \to a$. Then by proposition 6, $f(x_n) \to f(a)$ an so also $g(f(x_n)) \to g(f(a))$. So by proposition 6 $g \circ f$ continuous at $a$. \end{proof} \subsubsection*{Examples} \begin{enumerate}[(1)] \item $\RR \to \RR$ with usual metric. This is the same definition as when did it directly for $\RR$ only. So already know lots of continuous functions $\RR \to \RR$, for example polynomials, sin, exp, \dots \item Constant functions are continuous. Also if $X$ is any metric space and $f : X \to X$ by $f(x) = x$ for all $x \in X$ (the \emph{identity function on $X$}) then $f$ is continuous. \item Consider $\RR^n$ with the Euclidean metric and $\RR$ with the usual metric. \\ The \emph{projection maps} $\pi_i : \RR^n \to \RR$ given by $\pi_i (x) = x_i$ are continuous. (Why? We've seen convergence in $\RR^n$ of sequences then this is the same as convergence in each coordinate. Let's denote a sequence in $\RR^n$ by $(x^{(m)})_{m \ge 1}$. So for example, $x_5^{(3)}$ is the 5th coordinate of the 3rd term. We know $x^{(m)} \to x$ if and only if for each $x_i^{(m)} \to x_i$ i.e. for each $i$, $\pi_i(x^{(m)}) \to \pi_i(x)$. So by proposition 16 each $\pi_i$ is continuous.) Similarly, suppose $f_1, \dots, f_n : \RR \to \RR$. Let $f : \RR \to \RR^n$ be defined by $f(x) = (f_1(x), \dots, f_n(x))$. Then $f$ is continuous at a point if and only if all of $f_1, \dots, f_n$ are. \myskip Using these facts, and using example 1 and proposition 17 we have many continuous functions $\RR^n \to \RR^m$. For example consider $f : \RR^3 \to \RR^2$, \[ f(x, y, z) = (e^{-x} \sin y, 2x \cos z) \] is continuous. (Why? Take $W = (x, y, z) \in \RR^3$, we have $f_1(w) = e^{-\pi_1(w)} \sin \pi_2(w)$ and $f_2(w) = 2\pi_1(w) \cos \pi_3(w)$. So $f_1, f_2$ continuous so $f$ continuous.) \item Recall that if we have the Euclidean metric, the $l_1$ metric or the $l_\infty$ metric on $\RR^n$ then convergent sequences are same in each case. So by proposition 16, the continuous functions $X \to \RR^n$ or from $\RR^n \to Y$ are the same with each of these three metrics. \item Let $(X, d)$ be a discrete metric space and let $(Y, e)$ be any metric space. Which functions $f : X \to Y$ are continuous? Suppose $a \in X$ and $(x_n)$ a sequence in $X$ with $x_n \to a$. Then $(x_n)$ is eventually constant, i.e. for sufficiently large $n$, $x_n = a$ and so $f(x_n) = f(a)$. So $f(x_n) \to f(a)$. Hence \emph{every} function on a discrete metric space is continuous. \begin{hiddenflashcard}[cts-functions-on-discrete-metric-space] Which functions on a discrete metric space are continuous? \\ \cloze{All functions.} \end{hiddenflashcard} \end{enumerate} \subsection{Completeness} In section 1 we saw a version of general principle of convergence held in each of the three examples we considered. Does general principle of convergence hold in a general metric space? \begin{definition*}[Cauchy sequence] Let $(X, d)$ be a metric space and let $(x_n)$ be a sequence in $X$. We say $(x_n)$ is \emph{Cauchy} if \[ \forall \eps > 0 ~ \exists N ~ \forall m, n \ge N \quad d(x_m, x_n) < \eps \] \end{definition*} \noindent Exercise: $(x_n)$ convergent implies that $(x_n)$ Cauchy. \\ But the converse is not true in general. \begin{example*} Let $X = \RR \setminus \{0\}$ with the usual metric and let $x_n = \frac{1}{n}$. We saw previously that $(x_n)$ does not converge. Note that $X$ is a subspace of $\RR$. In $\RR$, $(x_n)$ is convergent ($x_n \to 0$) so $(x_n)$ is Cauchy in $\RR$ so $(x_n)$ is Cauchy in $X$. \end{example*} \begin{example*} Take $\QQ$ with the usual metric, and take a sequence $x_n \to \sqrt{2}$. Then $x_n$ is Cauchy but not convergent (in $\QQ$). \end{example*} \noindent This example with $\QQ$ is the main motivation for the following definition. \begin{flashcard} \begin{definition*} Let $(X, d)$ be a metric space. We say $X$ is \emph{complete} if \cloze{every Cauchy sequence in $X$ converges.} \end{definition*} \end{flashcard} \subsubsection*{Examples} \begin{enumerate}[(1)] \item Example above says $\RR \setminus \{0\}$ with usual metric is not complete. Similarly $\QQ$ with usual metric is not complete. \item General principle of convergence says $\RR$ with usual metric is complete. \par General principle of convergence for $\RR^n$ says $\RR^n$ with Euclidean metric is complete. \item General principle of uniform convergence (almost) says if $X \subset \RR$ and $B(X) = \{f : X \to \RR \mid \text{$f$ is bounded}\}$ with the uniform norm then $B(X)$ is complete. \end{enumerate}