Lecture 5. Limits and Continuity

It was already pointed out that convergence is at the heart of analysis as many problems can be solved by finding approximate solutions and proving that they converge. Consider an abstract equation such as $$ f(x)=0, $$ for a function $f:X\to \mathbb{R}$ and assume that we can find locations $x_n\in X$ such that $|f(x_n)|\leq 1/n$ for any $n\in \mathbb{N}$ (read approximate solutions). If $X$ carries a topology and the sequence $x_n$ lies in a compact subset $C$ of $X$, then it will have a convergent subsequence $(x_{n_k})_{k\in \mathbb{N}}$ with limit $x_\infty$. We would like to conclude that $x_\infty$ is a solution. For that we would need that $$ f(x_{n_k})\to 0\text{ as }n\to\infty. $$ This is unfortunately not true for arbotrary functions $f$. As this is apparently a desirable property for functions to have, it is given a name: continuity. Clearly as we don't know along which sequence we might be coming to $x_\infty$, the meaningful definition would have to be that $f$ is continuous at $x_\infty$ iff $$ f(x_n)\to f(x_\infty)\text{ whenever }x_n\to x_\infty, $$ or iff, given any $(x_n)_{n\in\mathbb{N}}$ with $x_n\to x_\infty\, (n\to\infty)$, $$ \forall\,\varepsilon>0\:\exists\, N\in \mathbb{N} \text{ s.t. }|f(x_n)-f(x_\infty)|\leq\varepsilon\text{ for } n\geq N. $$ This is precisely one of the definitions of continuity found in this lecture. Another practical motivation for considering continuity stems from the fact that many problems leading to a mathematical equation contain parameters $\lambda$, so that often equations for an unknown $x$ read $$ f(x,\lambda)=0. $$ As the parameters are often only known with a certain degree of accuracy, we cannot be certain of their exact value. Thus we would like to know whether a solution exist to the equation for $\lambda$ close to a fixed $\lambda _0$ provided one exists for $\lambda _0$ and whether these solutions are similar (close to each other). While a full understanding of this problem requires more than just continuity, you still get a sense that continuity will have to play some role, since it encodes the intuitive requirement that close enough arguments should lead to close values.

Definitions (Limit and Continuity)

Let $(X, d_X)$ and $(Y, d_Y)$ be two metric spaces. Given a function $f:E\to Y$ for a subset $E\subset X$, as well as $x_0\in E'\, ,\: y_0\in Y$ we say that
(i) $f$ is said to have $y_0$ as a limit as $x\to x_0$, or, in more mathematical notation $$ \lim_{x\to x_0} f(x)=y_0, $$ if and only if for any $\epsilon>0$, there is $\delta>0$ such that
$$ d_Y(f(x), y_0)\leq\epsilon\: \text{ whenever }\: 0 < d_X(x, x_0)\leq\delta \: \text{ and } \: x\in E. $$
(ii) $f$ is said to be continuous at $x_0\in E$ if and only if $\lim_{x\to x_0} f(x)=f(x_0)$
(iii) $f$ is called discontinuous at $x_0$ iff $f$ is not continuous at $x_0$.

Theorem

Let $f, g:\mathbb{R}^n\to \mathbb{R}$ be two functions and $x_0\in \mathbb{R}^n$. Then
(i) $\lim_{x\to x_0}\bigl[ f(x)\pm g(x)\bigr]=\lim_{x\to x_0} f(x) \pm \lim_{x\to x_0} g(x)$,
(ii) $\lim_{x\to x_0}\bigl(f(x)g(x)\bigr)=\lim_{x\to x_0} f(x) \cdot\lim_{x\to x_0} g(x)$,
(iii) $\lim_{x\to x_0} \big(f(x) g(x)^{-1}\big)=\lim_{x\to x_0} f(x)\bigl[\lim_{x\to x_0} g(x)\bigr]^{-1}$, whenever $\lim_{x\to x_0} g(x)\ne 0$,
provided that all limits exist.

Example

If $(x_0, y_0) \ne (0,0)$, that is, if $x_0^2 + y_0^2\neq 0$, then $(x,y)\ne (0,0)$ for $(x,y)$ sufficiently close to $(x_0,y_0)$. It follows that $$ \lim_{(x,y) \to (x_0, y_0)} f(x,y) = \lim_{x \to x_0, y \to y_0} \dfrac{xy}{x^2+y^2} = \dfrac{x_0y_0}{x_0^2 + y_0^2}=f(x_0, y_0)\, . $$ If, on the other hand, $(x_0, y_0) = (0,0)$, then $$ \lim_{x=y\to 0} f(x, y) = \lim_{x\to 0}\frac{x x}{x^2+x^2}=\frac{1}{2} \quad\text{ and }\: \lim_{x \to 0, y=0} f(x, y)=\lim_{x\to 0}\frac{0}{x^2}=0. $$ We conclude that $\lim_{(x, y)\to (0,0)} f(x, y)$ does not exist and thus $f$ is continuous on $\mathbb{R}\setminus\{(0,0)\}$.

Example

Example

We claim that $C = \mathbb{R}^2 \setminus \{(0,0)\}$. For any $(x_0, y_0) \in C$, we have that \begin{align*} \lim_{(x,y)\to(x_0,y_0)} f(x,y) & =\lim_{(x,y)\to(x_0,y_0)} \dfrac{xy^2}{x^2 + y^4}=\dfrac{x_0y_0^2}{x_0^2 + y_0^4} = f(x_0, y_0), \end{align*} so that $f$ is continuous at $(x_0, y_0)$. Notice, however, that $$ \lim_{x\to0, y= 0} \dfrac{xy^2}{x^2+ y^4} = \lim_{x\to 0} \dfrac{0}{x^2} = 0, $$ and that $$ \lim_{x=y^2 \to0} \dfrac{xy^2}{x^2+ y^4} = \lim_{y\to 0} \dfrac{y^2y^2}{y^4 + y^4}=1/2. $$ Thus $\lim_{(x,y) \to (0,0)} f(x,y) = \lim_{(x,y) \to (0,0)} \dfrac{xy^2}{x^2+y^4}$ does not exist, concluding the proof.

Example

The Dirichlet function \[D(x)= \left\{ \begin{array}{l l} 1 & \quad \mbox{$x$ is rational}\\ 0 & \quad \mbox{$x$ irrational}\\ \end{array} \right. \] is discontinuous at every point of $\mathbb{R}$ due to the fact that $\mathbb{Q}$ and $\mathbb{R}\setminus{Q}$ both are dense in $\mathbb{R} $.

Example

The Riemann function $$ R(x)=\begin{cases} 0\, ,&\text{if }x\text{ is irrational}\\ 1/n\, ,&\text{if }x=m/n\, ,\: (m,n)=1\\ 1\, , &\text{if } x=0,1 \end{cases} $$ is such a function. Indeed, if $x_0 \in (0,1)$ is an irrational number, we know that $R(x_0) = 0$. Let $\epsilon>0$ and consider the set $$ E_{\epsilon}=:\{m/n\in (0,1)\, |\, (m,n) =1\text{ and }\dfrac{1}{n}\ge\epsilon\} $$ It is easily seen that $E_{\epsilon}$ is a finite set, so that $$ E_{\epsilon}=\{{p_j /q_j}\, |\, j=1,\dots, k\}\, . $$ Then $$ R(x)|\leq\epsilon\text{ whenever }|x-x_0|\leq\delta $$ for $\delta=\min\{|p_j/q_j-x_0|\, :\, j=1,\dots, k\}$. It follows that $R(x)$ is continuous at $x_0$. If, on the other hand, $x_0$ is rational, then $R(x_0)>0$, while, by density of the irrationals, there are inftinitely many irrational numbers $x$ in any interval around $x_0$ where $R(x)=0$.

Definition (monotony)

A function $f:(a,b)\to \mathbb{R}$ is called [strictly] increasing if $$ f(x_1) \le f(x_2)\: [f(x_1) < f(x_2)],\text{ whenever }x_1\leq x_2. $$

Theorem

Let $f:(a,b)\to \mathbb{R}$ be an increasing function and set $$ D_f = \{x_0 \in (a,b)\, |\, f\text{ is discontinuous at }x_0 \}. $$ Then, $D_f$ is at most countable.

Let $x_0 \in D_f$. Since $f$ is increasing, we have that $$ \lim_{x \to x_0\pm} f(x) = f(x_0\pm) $$ exist. For the same reason we must also have that $f(x_0-)< f(x_0+)$ at points of discontinuity. Define $$ I_{x_0} = \bigl(f(x_0-), f(x_0+)\bigr). $$ We also have that $f(x_0+)\le f(x_1-)$ for $x_0 < x_1$ and that $$ I_{x_0} \cap I_{x_1} =\emptyset,\: x_0\ne x_1, $$ for $x_0, x_1 \in D_f$. For $x\in D_f$ choose a rational number $\gamma_{x} \in I_{x}$. It follows that $$ \gamma_{x_0} < \gamma_{x_1}\text{ if }x_0,x_1 \in D_f\text{ satisfty }x_0 < x_1. $$ The proof is completed by observing that $\{\gamma_{x_0} : x_0 \in D_f \}$ is at most countable as a subset of $\mathbb{Q}$ and so is $D_f$.

Example

It follows from $f(1)=f(0)\, f(1)$ that $f(0)=1$. The fact that $f(0)=f(-1)\, f(1)$ yields $f(-1)=e^{-1}$. Thus $$ f(m) = f(1) f(m-1) = f(1)^m= e^m\, ,\: m \in \mathbb{Z}. $$ Now $f(1)=f(n/n)=f(1/n)^n$ implies that $f(1/n)=e^{1/n}$ and that $f(m/n)=f(1/n)^m=e^{m/n}$. Finally continuity combined with the density of $\mathbb{Q}$ in $\mathbb{R}$ yields $$ f(x)=\lim_{n\to\infty} f(r_n)=\lim_{n\to\infty} e^{r_n}=e^x, $$ by choosing a sequence $\{r_n\}$ in $\mathbb{Q}$ such that $x=\lim_{n\to\infty}r_n$.

Example

Definition (Convexity)

Let $E$ be convex and $f:E\to \mathbb{R}$ be a function. We say that $f$ is convex iff $$ f\bigl(\lambda x+(1-\lambda) y\bigr)\le \lambda f(x)+(1-\lambda) f(y),\: x,y\in E,\:\lambda\in (0,1)\, . $$

Exercise

Theorem

If $f:E\to \mathbb{R}$ is a convex function on a convex subset $E\subset\mathbb{R}^n$, then $f$ is locally bounded and continuous on $E$.

We consider $n=1$ case, so that $E$ is an interval. If $E$ has no interior, there is nothing to prove. Otherwise take any $a,b, c\in E$ such that $a < b < c$. Using $$ f\bigl( \lambda a+(1-\lambda)b\bigr)\leq\lambda f(a)+(1-\lambda)f(b), $$ it follows that $$ \limsup_{x\to b-} f(x)=\limsup_{x\to b-} f(\lambda_x a+(1-\lambda_x)b)\le \limsup_{x\to b-} \bigl[\lambda_x\, f(a)+(1-\lambda_x) f(b)\bigr]\le f(b), $$ since $\lambda_x\to 0$ if $x\to b$. The validity of $$ f\bigl( \lambda b+(1-\lambda)c\bigr)\leq\lambda f(b)+(1-\lambda)f(c) $$ implies that $$ \limsup_{x\to b+}f(x)=\limsup_{\lambda\to 1-} f(\lambda b+(1-\lambda)c)\le f(b). $$ Now, for any $0 < \epsilon < \min \{b-a, c-b\}$, one has $$ f(b)=f({b-\epsilon\over 2}+{b+\epsilon\over 2})\le {1\over 2}f(b-\epsilon)+{1\over 2} f(b+\epsilon). $$ This, in turn, implies that $$ f(b)\le {1\over 2} \limsup_{\epsilon\to 0+}f(b-\epsilon)+{1\over 2}\liminf_{\epsilon\to 0+} f(b+\epsilon)\le {1\over 2} f(b)+{1\over 2} \liminf_{\epsilon\to 0+} f(b+\epsilon) $$ Thus $$ f(b)\le \liminf_{x\to b+}f(x)\text{ and similarly }f(b)\le \liminf_{x\to b-}f(x). $$ and therefore $$ \limsup_{x\to b+}f(x)=f(b)=\liminf_{x\to b+}f(x)\hbox{ and } \limsup_{x\to b-}f(x)=f(b)=\liminf_{x\to b-}f(x). $$ We conclude that $f$ is continuous at $b$, and hence continuous in $E$. Therefore, it is locally bounded on $E$.

Exercise

Properties of continuous functions

Theorem

Let $(X, d)$ be a metric space and the function $f:X\to \mathbb{R}$ be continuous at $x_0 \in X $. If $f(x_0)>0$, then there is $\delta>0$ such that $$ f(x)\ge \frac{f(x_0)}{2}>0\text{ for all }x\in \mathbb{B}(x_0, \delta). $$

Setting $\epsilon=\frac{f(x_0)}{2}>0$ continuity yields a $\delta>0$ such that $$ |f(x)-f(x_0)|\leq\epsilon=\frac{f(x_0)}{2}\text{ for }x\in \mathbb{B}(x_0,\delta). $$ Thus $f(x)-f(x_0)\geq-\frac{f(x_0)}{2}$ and thus $f(x)\geq \frac{f(x_0)}{2}>0$ for any $x\in \mathbb{B}(x_0, \delta)$.

Theorem (Topological Characterization)

Let $(X, d_X)$ and $(Y, d_Y)$ be two metric spaces. Then $f: X \to Y$ is continuous if and only if preimages of open sets are open, i.e. iff $$ f^{-1}(V) = \big\{ x\in X\, |\, f(x)\in V\big\}\text{ is open in }X\text{ whenever }V\text{ is open in }Y\, . $$

"$\Rightarrow$" Assume $f$ is continuous from $X \to Y$ and let $V\subset Y$ be open. We need to prove $f^{-1}(V)$ is open in $X$. Take $x_0\in f^{-1}(V)$ so that $y_0=f(x_0)\in V$. Since $V$ is open, there is $\epsilon>0$ with $$ \mathbb{B}_Y(y_0,\epsilon)\subset V. $$ Continuity of $f$ at $x_0$ yields $\delta>0$ such that $$ f\bigl(\mathbb{B}_X(x_0, \delta)\bigr)\subset \mathbb{B}_Y(y_0,\epsilon)\subset V, $$ so that $\mathbb{B}(x_0, \delta)\subset f^{-1}(V)$ and $f^{-1}(V)$ is open since $x_0$ was arbitrary.
"$\Leftarrow$" Assume $f^{-1}(V)$ is open in $X$ whenever $V$ is open in $Y$. We show $f$ is continuous. For any $x_0 \in X$, $f(x_0)\in Y$ and, for any $\epsilon>0$, $\mathbb{B}_Y(f(x_0), \epsilon)$ is an open set in $Y$. Then $$ f^{-1}\Bigl(\mathbb{B}_Y\bigl(f(x_0), \epsilon\bigr)\Bigr)\text{ is open in }X, $$ and contains $x_0$. So, there must be $\delta>0$ with $$ \mathbb{B}_X(x_0,\delta)\subset f^{-1}\Bigl(\mathbb{B}_Y\bigl(f(x_0), \epsilon\bigr)\Bigr) $$ or $f\bigl(\mathbb{B}_X(x_0,\delta)\bigr)\subset \mathbb{B}_Y\bigl(f(x_0),\epsilon\bigr)$. Thus, $f$ is continuous at $x_0$.

Exercise

Theorem (Extrema)

Let $(X, d)$ be a complete metric space and let $K\subset X$ be compact. Let $F: K \to \mathbb{R}$ be a continuous function. Then there are $x_0, y_0 \in K$ such that $$ f(x_0)\le f(x)\le f(y_0),\: x\in K. $$

Let $$ M=\sup\big\{f(x): x\in K\big\} \text{ and } m=\inf\big\{f(x): x\in K\big\}. $$ By definition, there are sequences $\{x_n\}_{n=}^\infty$ and $\{y_n\}_{n=1}^\infty$ in $K$ such that $$ M=\lim_{n\to\infty} f(y_n)\quad\hbox{ and } \lim_{n\to \infty} f(x_n)=m. $$ Since $K$ is compact and $X$ is complete, there are subsequences $(x_{n_k})_{k\in\mathbb{N}}$ and $(y_{n_k})_{k\in \mathbb{N} }$ as well as $x_0, y_0\in K$ such that $$ \lim_{k\to\infty} x_{n_k}=x_0\quad\hbox{and }\lim_{k\to\infty} y_{n_k}=y_0. $$ By continuity of $f$ at $x_0$ and $y_0$ it follows that $$ f(x_0)=\lim_{k\to\infty} f(x_{n_k})=m\text{ and } f(y_0)=\lim_{k\to\infty}f(y_{n_k})=M, $$ and the proof is complete.

Theorem (Intermediate Value Theorem)

Let $f:[a, b]\to \mathbb{R}$ be continuous. Then $$ f\bigl([a,b]\bigr)=[m, M] $$ where $M=\max\big\{f(x)\, |\, x\in K\big\}$ and $m=min\big\{f(x)\, |\, x\in K\big\}$.

Exercise

Uniform continuity

Definition

Let $(X, d_X)$ and $(Y,d_Y)$ be metric spaces and $K\subset X$. $f: K \to Y$ is said to be uniformly continuous on $K$ iff, for any $\epsilon>0$, there is $\delta>0$ such that $$ d_Y\bigl(f(x),f(y)\bigr)\leq\epsilon\text{ whenever }x,y \in K\text{ with }d_X(x,y)\leq\delta. $$

Proposition

If $f$ is uniformly continuous on $K$, then $f:K\to Y$ is continuous on $K$. The converse is not true.

Nexample

Let $\epsilon_0=1$ and any $\delta>0$. Take $x_\delta = 1/\delta$, $y_\delta = 1/\delta + \delta/2$ so that $$ |x_\delta - y_\delta| = \delta/2 < \delta $$ but \begin{equation*} |f(x_\delta) - f(y_\delta)| = |x_\delta + y_\delta||x_\delta-y_\delta| = (1/\delta + 1/\delta + \delta) \delta/2\ge 2/\delta \cdot \delta/2 = 1 =\epsilon_0\, . \end{equation*}

Theorem

Let $(X, d)$ be a complete metric space and $K\subset X$ a compact set. If $f:K\to Y$ is continuous, then $f$ is uniformly continuous.

If $f$ is not uniformly continuous on $K$, then we can find $\epsilon_0>0$ such that, for any $\delta>0$, there are $x_\delta$, $y_\delta \in K$ with $$ d_X(x_{\delta}, y_{\delta})\leq\delta \text{ and }d_Y\bigl(f(x_\delta), f(y_\delta)\bigr)\ge\epsilon_0 \, . $$ For $n\in\mathbb{N}$, choose $\delta=1/n$ and $x_n, y_n \in K$ such that $$ d_X(x_n, y_n)\leq 1/n \text{ and }d_Y\bigl(f(x_n), f(y_n)\bigr)\ge\epsilon_0 $$ Since $K$ is compact, there are subsequences $(x_{n_k})_{k\in \mathbb{N}}$ and $(y_{n_k})_{k\in \mathbb{N}}$ with $$ \lim_{k\to \infty} x_{n_k}=x=\lim_{k\to \infty} y_{n_k}\, , $$ for some $x\in K$. Thus \begin{equation*} 0<\epsilon_0\leq d_Y\bigl(f(x_{n_k}), f(y_{n_k})\bigr)\leq d_Y\bigl(f(x_{n_k}), f(x)\bigr)+ d_Y\bigl(f(x), f(y_{n_k})\bigr)\, , \end{equation*} which contradicts the continuity of $f$ at $x$.

Definition

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and $f:X\to Y$ be given. The function $f$ is called Hölder continuous of exponent $\alpha\in(0,1)$ if there is a constant $c\geq 0$ such that $$ d_Y\bigl(f(x),f(y)\bigr)\leq c\, d_X(x,y)^\alpha\, ,\: x,y\in X\, . $$ If the inequality holds with $\alpha=1$, the function is called Lipschitz continuous continuous.

Example

Example

Example

Method 1: Fix $\epsilon>0$. Since $f$ is continuous on the compact interval $[0,2]$, it is uniformly continuous there and we can find $\delta_1>0$ such that $$ |f(x_1)-f(x_2)|\leq\epsilon\text{ if }x_1, x_2\in [0,2]\text{ and }|x_1-x_2|<\delta_1. $$ Let $\delta_2=2\epsilon$. Since $f'(x) = \dfrac{1}{2} \dfrac{1}{\sqrt{x}}$ and $|f'(x)| \le \dfrac{1}{2}$ on $[1,\infty)$, the Mean Value Theorem (proved later) yields $$ |f(x_1) - f(x_2) | = f'(\xi)|x_1 - x_2| \le \dfrac{1}{2} |x_1 - x_2|\leq\epsilon $$ for $x_1, x_2 \in [1, \infty)$ such that $|x_1-x_2|\leq\delta_2$, and some $\xi\in(x_1,x_2)$. Finally choosing $$ \delta=\min\{\delta_1, \delta_2, 1\}, $$ it is easy to see that $x_1, x_2\in (0,2]$ and $|x_1-x_2|<\delta_1$ or $x_1, x_2\in[1,\infty)$ and $|x_1-x_2|<\delta_2$ whenever $x_1, x_2\in (0,\infty)$ and $|x_1-x_2|<\delta$. Therefore, in any case, we have that $$ |f(x_1)-f(x_2)|\leq\epsilon\text{ for }x_1,x_2\in(0,\infty)\text{ with }|x_1, x_2|\leq\delta, $$ as desired.
Method 2: Set $\delta=\epsilon^2$ for $\epsilon>0$. For any $0 < x_1\le x_2 <\infty$ and $x_2-x_1<\delta$, one either has that $ x_2\le \epsilon^2$, in which case $$ |\sqrt{x_2}-\sqrt{x_1}|\le \sqrt{x_2}\le \epsilon $$ or $x_2\ge \epsilon^2$ and then $$ |\sqrt{x_2}-\sqrt{x_1}|= {x_2-x_1\over \sqrt{x_2}+\sqrt{x_1}}\le {\epsilon^2\over \epsilon}=\epsilon. $$ Thus $\sqrt{x}$ is uniformly continuous on $(0,\infty)$.

Exercise

Example

Since $\frac{\partial f}{\partial x_j} (x) = \frac{2 x_j}{1+|x|^2}$, for any $x\in \mathbb{R}^n$, it follows that $$ |\nabla f(x)|=\sum_{j=1}^n \big | \frac{\partial f}{\partial x_j}\big | ^2=\sum_{n=1}^n\frac{4 x_j^2}{(1+ | x| ^2)^2}= \frac{4 | x| ^2 }{(1+ |x|^2)^2}= \bigl(\frac{2 |x|}{(1+|x| ^2)}\bigr)^2 \le 1. $$ Then we have that $f$ is uniformly continuous on $\mathbb{R}^n$. In fact \begin{equation*} |f(x) - f(y)|=|\nabla f(\xi) \cdot (x-y) | \le |\nabla f(\xi) | | x-y|\le | x-y| \end{equation*} for $\xi=\tau x +(1-\tau)y$ and some $\tau\in(0,1)$.

Theorem (Extension of Uniformly Continuous Functions)

Let $(X,d)$ be a metric space and $K \subset X$. If $f:K\to \mathbb{R}$ is uniformly continuous, then it can be extended to a uniformly continuous function $\tilde f$ to $\overline{K}$.

Fix $x_0 \in K' \setminus K$. How can we define $\tilde f(x_0)$? Choose first a sequence $\{x_n\}_{n=1}^\infty$ in $K$ with $\lim _{n\to\infty}x_n=x_0$. Uniform continuity ensures that $\bigl( f(x_n)\bigr)_{n\in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$ (verify this for yourself!). Then take $\tilde f(x_0)$ to be its limit, which exists by completeness of $\mathbb{R}$.
Next we need to make sure that this limit does not depend on the choice of approximating sequence for $x_0$. Let $\{y_n\}_{n=1}^\infty$ be any other sequence in $K$ converging to $x_0$. Then $$ |f(y_n)-\tilde f(x_0)|\le |\tilde f(x_0)-f(x_n)|+|f(x_n)-f(y_n)|\, ,\: n\in\mathbb{N} $$ and uniform continuity ensures smallness of the second term on the righ-hand-side of the inequality. Thus $f$ can be extended uniquely to $\tilde f:\overline{K}\to\mathbb{R}$.
It remains to show that $\tilde f$ is uniformly continuous on $\overline{K}$. Given $\epsilon>0$, uniform continuity of $f$ yields a $\delta>0$ such that $$ |f(x)-f(y)|\leq\epsilon\text{ whenever }x, y\in K\text{ with }d(x, y)\leq\delta. $$ For any $x, y\in \overline{K}$ with $d(x, y)<\delta/3$ there are $x', y'\in K$ such that $$ d(x, x')\leq\delta/3,\: |\tilde f(x)-f(x')|\leq\epsilon\text{ and that }d(y, y')\leq\delta/3,\: |\tilde f(y)-f(y')|\leq\epsilon. $$ It follows that $$ d(x', y')\le d(x', x)+d(x,y)+d(y, y')\leq\delta $$ and that $$ |\tilde f(x)-\tilde f(y)|\le|\tilde f(x)-f(x')|+|f(x')-f(y')|+|f(y')-\tilde f(y)|\leq\epsilon +\epsilon+\epsilon=3\epsilon, $$ and $f$ is indeed uniformly continuous on $\bar{K}$.

Example

We have that $\mathbb{\bar{Q}} = \mathbb{R}$, since $\mathbb{Q}$ is dense in $\mathbb{R}$. Thus existence of an extension now follows from the previous theorem.

Continuity of the Inverse

Definition (Inverse)

Let $f: X \to Y$ be one-to-one and onto. The inverse function $f^{-1}: Y \to X$ is defined by setting $$ f^{-1} (y) = x \text{ for }x\in X\text{ such that }f(x) = y. $$

Lemma

If $f:X\to Y$ is continuous and $K \subset X$ is compact, then so is $f(K)$.

Any open covering $\{V_\lambda: \lambda \in I\}$ of $f(K)$ yields the open cover $\{f^{-1} (V_\lambda): \lambda \in I\}$ for $K$ by continuity of $f$. Now $K$ is compact and there must be a finite subcover, that is, $\{ f^{-1} (V_{\lambda_j})\, |\, j=1,\dots,n\}$ with $$ K\subset \cup_{j=1}^n f^{-1} (V_{\lambda_n})\, . $$ It follows that $f(K) \subset \cup_{j=1}^n V_{\lambda_j}$ and $f(K)$ is compact.

Exercise

Theorem

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and $X$ be compact. If $f:X \to Y$ is a continuous, one-to-one, and onto function, then $f^{-1}: Y\to X$ is also continuous.

Let $U$ be any open set in $X$. Then $U^c$ is closed in $X$. Since $X$ is compact, so is $U^c$. Then $f(U^c)$ is compact in $Y$ and therefore also closed. Now, since $f$ is injective, $$ f(U) = Y\setminus f(U^c) $$ is open and $f^{-1}: Y\to X$ is continuous.