Continuous functions-(2)

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Lemma 1

Let $f:X\to Y$ be a function and let $A\subset X, B\subset Y$. Then, \(\begin{align} \begin{split} f^{-1}(f(A)) &\supset A \\ f(f^{-1}(B)) &\subset B \end{split} \end{align} \label{set}\)


Let $x \in A$ be given. Then $f(x)\in f(A)$. Since \(f^{-1}(f(A)) = \{x\in X: f(x)\in A \}\), $x\inf^{-1}(f(A))$.

$\therefore f^{-1}(f(A)) \supset A$.

Now, let $y\in f(f^{-1}(B))$ be given. Then $y=f(x_0)$ for some $x_0 \in f^{-1}(B)$. Since $x_0 \in f^{-1}(B), y= f(x_0)\in B$.
$\therefore f(f^{-1}(B))\subset B$ \(\tag*{$\square$}\)

Theorem 4.2.8

Let $K$ is a compact set and let $f:K\to \mathbb{R}$ be a continuous function on $K$. Then $f(K)$ is compact set.

<proof> Let \(\{V_\alpha\}_{\alpha \in \Lambda}\) be an open cover of $f(K)$. Since $f$ is continuous on $K, f^{-1}(V_\alpha)$ is open in $K$ for all $\alpha \in \Lambda$. Then there exists an open set $U_\alpha$ in $\mathbb{R}$ such that $f^{-1}(V_\alpha) = U_\alpha \cap K$.

Now we want to show that $K\subset \cup_{\alpha\in \Lambda} U_\alpha$. \(\begin{align} \begin{split} p\in K &\Rightarrow f(p) \in f(K) \\ &\Rightarrow f(p) \in \cup_{\alpha \in \Lambda}U_\alpha\\ &\Rightarrow f(p) \in U_\alpha \text{ for some } \alpha \in \Lambda \\ &\Rightarrow p\in f^{-1}(V_\alpha) = U_\alpha \cap K \subset U_\alpha \end{split} \end{align}\) So, \(\{U_\alpha \}_{\alpha \in \Lambda}\) is an open cover of $K$. Since $K$ is compact, there are $\alpha_1,\ldots, \alpha_n \in \Lambda$ such that $K\subset \cup_{i=1}^n U_{\alpha_i}$. In other words, \(\begin{align} \begin{split} K&\subset (\cup_{i=1}^n U_{\alpha_i}) \cap K \\ &=\cup_{i=1}^n(U_{\alpha_i}\cap K) \\ &= \cup_{i=1}^n f^{-1}(V_{\alpha_i}) \end{split} \end{align}\) Thus, $f(K)\subset f(\cup_{i=1}^n f^{-1}(V_{\alpha_i}))=\cup_{i=1}^n f(f^{-1}(V_{\alpha_i}))$. Since $f(f^{-1}(V_{\alpha_i})) \subset V_{\alpha_i}$ by Eq.\ref{set}, $\cup_{i=1}^n f(f^{-1}(V_{\alpha_i})) \subset \cup_{i=1}^n V_{\alpha_i}$.

$\therefore f(K)$ is compact.


Lemma 2

Let $A$ be a closed set. Then $\sup A \in A$.

<proof> For any $n\in \mathbb{N}$, there is $x_n \in A$ such that $\sup A -1/n \leq x_n \leq \sup A$. If there is $n_0\in\mathbb{N}$ such that $x_{n_0} = \sup A$, then $\sup A \in A$.
Suppose that $x_n \neq \sup A$ for all $n\in\mathbb{N}$. Then $\sup A \in A^\prime$. Since $A$ is closed, $A^\prime \subset A$.

$\therefore \sup A \in A$. \(\tag*{$\square$}\)

Corollary 4.2.9

Let $K$ be a compact subset of $\mathbb{R}$ and let $f:K\to\mathbb{R}$ be a continuous function. Then there exists $p,q \in K$ such that \(\begin{align} f(q) \leq f(x) \leq f(p) \text{ for all }x\in K \end{align}\)

<proof> By the above theorem, $f(K)$ is compact. By the Heine-Borel-Bolzano-Weierstrass theorem, $f(K)$ is closed and bounded. Let \(\begin{align} M:= \sup \{f(x): x\in K\} \end{align}\) Since $f(K)$ is bounded, $M < +\infty$. Since $f(K)$ is closed, $M\in f(K)$. So, there is $p\in K$ such that $f(p) = M$, i.e. $f(x) \leq f(p)$ for all $x\in K$. Similarly for \(m=\inf\{f(x):x\in K\}\).


Theorem 4.2.10

Let $X,Y$ be metric spaces and let $f:X\to Y$ be a continuous function. If $E\subset X$ is connected, then $f(E)$ is connected.

<Proof> Proof by contraposition. Suppose that $f(E)$ is disconnected, i.e., $f(E)=A\cup B$ where $\overline{A}\cup B =\emptyset$ and $A\cup \overline{B}=\emptyset$. Consider,

\[\begin{align*} G&=f^{-1}(A)\cap E\\ H&=f^{-1}(B)\cap E \end{align*}\]


\[\begin{align*} (f^{-1}(A)\cap E )\cup (f^{-1}(B)\cap E)&=E\cap (f^{-1}(A)\cup f^{-1}(B))\\ &=E\cap (f^{-1}(A\cup B))\\ &=f^{-1}(A)\cup f^{-1}(B), \end{align*}\]

$E=G\cup H.$ By definition of $G$,

\[\begin{align*} G=f^{-1}(A)\cap E \subset f^{-1}(A) \subset f^{-1}(\overline{A}). \end{align*}\]

Since $f$ is continuous, $f^{-1}(\overline{A}^c)$ is open in $X$. We know that $f^{-1}(S^c)=(f^{-1}(S))^{-1}$. Thus, $f^{-1}(\overline{A}^c)$ is closed, which implies that $\overline{G}\subset f^{-1}(\overline{A}) (\because S_1 \subset S_2 \Rightarrow \overline{S}_1\subset \overline{S}_2).$ Similarly, $H\subset f^{-1}(B)$.

Then, we get the following equation \(\begin{align*} \overline{G}\cap H &= f^{-1}\left(\overline{A}\right)\cap f^{-1}(B) \\ &=f^{-1}\left(\overline{A}\cap B\right) \\ &=\emptyset \end{align*}\)

Similarly, we get $G\cap \overline{H}=\emptyset$. Thus, $E$ is disconnected. By contraposition, we show that $f(E)$ is connected.


Theorem 4.2.11 (Intermediate Value Theorem)

Let $f: [a,b]\to\mathbb{R}$ be a continuous function with $f(a) < f(b)$.
If $\gamma \in\mathbb{R}$ s.t. $f(a)<\gamma <f(b)$, then there is $c\in (a,b)$ s.t. $f(c) =\gamma$.

<proof> Let \(\begin{align} A :=\{x\in [a,b]: f(x)\leq \gamma \} \end{align}\) Then $A \neq \emptyset$ and is bounded above by $b$. By the least upper bound property, there is a supremum $c=\sup A$. Then $c\leq b$.

Now, we want to show that $f(c) = \gamma$. Since $c=\sup A$, $c\in A$ or $c\in A^\prime$. If $c\in A$, then $f(c)\leq \gamma$.

Suppose that $c\in A^\prime$. Then there exists \(\{x_n\}\subset A\) such that $x_n\to c$ as $n\to\infty$ with $x_n\neq c$. Since $x_n \in A, f(x_n)\leq \gamma$ for all $n\in \mathbb{N}$. Since $f$ is continuous and $c$ is a limit point of $[a,b],$ \(\begin{align} f(c) = \lim_{n\to\infty}f(x_n) \leq \gamma \end{align}\) Thus, $f(c) \leq \gamma$.

Suppose that $f(c) \lneq \gamma$. Take \(\begin{align} \epsilon_0 := \frac{1}{2}(\gamma -f(c)) > 0 \end{align}\) Since $f$ is continuous at $c$, there is $\delta >0$ such that \(\begin{align} x \in N_\delta (c) \cap [a,b] \Rightarrow f(c) -\epsilon_0 < f(x) < f(c) + \epsilon_0 \end{align}\) Since $f(c) \lneq \gamma, c\neq b$. Then $(c,b]\cap N_\delta (c) \neq \emptyset$. If $x\in (c,b]$ with $c<x<c+\delta$, then \(\begin{align} \begin{split} f(x) < f(c) + \epsilon_0 &= f(c) + \frac{1}{2}(\gamma - f(c)) \\ &=\frac{1}{2}(\gamma + f(c)) \\ &<\gamma \end{split} \end{align}\) Since $x >c =\sup A, x \not\in A$ That is $f(x) > \gamma$. But it is a contradiction because $c$ is supremum of $A$.

$\therefore f(c) = \gamma$


Another proof as a corollary to Theorem 4.2.10.

<Proof> Since $[a,b]$ is connected, $f([a,b])$ is connected. By Theorem 3.1.18, $(f(a), f(b))\subset f([a,b])=\text{im}f$. Thus, for any $c\in (f(a), f(b)),$ there is some $x\in [a,b]$ such that $c=f(x)$. Since $c\neq f(a)$ and $c\neq f(b)$, $x\in (a,b)$.


Corollary 4.2.12

Let $I \subset \mathbb{R}$ be an interval and let $f:I\to \mathbb{R}$ be a continuous function. Then $f(I)$ is an interval.

<proof> Let $s,t \in f(I)$ with $s<t$ such that $f(a) = s, f(b) = t$.
Let $s <\gamma <t$ be given. If $a<b$, then since $f$ is continuous on $I$, by the intermediate value theorem there exists $c\in (a,b)$ such that $f(c) = \gamma$.
$\therefore \gamma \in f(I)$
A similar argument also holds if $a>b$.


Corollary 4.2.13

For every $\gamma >0$ and for every $n\in\mathbb{N}$, there exists a unique $y>0$ such that $y^n = \gamma$.

<proof> Let $f(x) := x^n$. Since $f$ is continuous on $\mathbb{R}$, it is also continuous on $[0,\gamma+1]$. Since $(\gamma +1)^n > \gamma, 0<\gamma <(\gamma+1)^n$. By the mean value theorem, there exists $y\in\mathbb{R}$ such that $f(y) = y^n=\gamma$.

Now we want to show the uniqueness of $y$. Suppose that $y^n_1 = y^n_2=\gamma$. Then, \((y^n_1 -y^n_2 )= (y_1 - y_2)(y^{n-1}_1 + y^{n-2}_1y_2 + \cdots + y_1y^{n-1}_2)=0\) Since $y_1, y_2 >0, y_1 = y_2$.

Corollary 4.2.14

If $f:[0,1]\to[0,1]$ is continuous, then there is $y\in [0,1]$ such that $f(y) = y$.

<proof> Let $g(x) := f(x) = x$. Then $g(0) = f(0) \geq 0$, $g(1) = f(1)-1 \leq 0$.
If $g(0) =0$ or $g(1) =0$, then $f(0) =0$ or $f(1)=1$.

Suppose that $g(0) \neq 0$ and $g(1) \neq 0$. Since $f$ is continuous, by the mean value theorem, there is $y \in \mathbb{R}$ such that $g(y)=0 (\because g(1) < 0 < g(0) )$.

$\therefore \exists y\in\mathbb{R}$ such that $f(y) = y$.



  • Manfred Stoll, Introduction to Real Analysis, Pearson