Proof of Intermediate Value Theorem on a Closed Real Interval, and Continuous Images of Intervals

Throughout, $d_{\mathbb{R}}(s,t)=|s-t|$ for $s,t\in\mathbb{R}$, by The Distance on the Real Numbers (i); for $a\le b$, $t\in[a,b]$ if and only if $t\in\mathbb{R}$ and $a\le t\le b$, by Intervals of the Real Line (ii); and for $u,v\in\mathbb{R}$, Rules of Arithmetic and Order in an Ordered Field (x) gives $-|u|\le u\le|u|$, $|-u|=|u|$, and $|u|\le v$ if and only if $-v\le u\le v$. In particular, if $0\le u$ then $-u\le0\le u$ by Rules of Arithmetic and Order in an Ordered Field (v), and so $|u|\le u$.

Proof of (i).

Case 1: $f(a)\le y\le f(b)$. This case is proved for every map $f:[a,b]\to\mathbb{R}$ continuous on $[a,b]$ and every such $y$. Let $S=\{x\in[a,b]:f(x)\le y\}$ (Typed Terms under a Predicative Guard: Variables, Parameters, Values of Maps and Ordered Pairs, and Predicative Equations and Memberships between Them (vii)), a subset of $\mathbb{R}$. As $a\le a\le b$ and $f(a)\le y$, $a\in S$, so $S\neq\emptyset$; and $b$ is an upper bound of $S$, so $S$ is bounded above. By The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (vii), $S$ has a supremum, unique by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv); let $c=\sup S$. By Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv) and Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iii), $c$ is an upper bound of $S$, and $c\le u$ for every upper bound $u$ of $S$. Hence $a\le c$, as $a\in S$, and $c\le b$, as $b$ is an upper bound of $S$; so $c\in[a,b]$. By Continuous Map Between Metric Spaces (ii), $f$ is continuous at $c$ relative to $[a,b]$, so by Continuous Map Between Metric Spaces (i), for every real $\varepsilon>0$ there is a real $\delta>0$ such that every $x\in[a,b]$ with $|c-x|<\delta$ satisfies $|f(x)-f(c)|<\varepsilon$. In each of Steps 1 and 2, $\varepsilon$ is fixed first, from $c$ and $y$; then $\delta$ is taken for this $\varepsilon$; and then the points tested against $\delta$.

Step 1: $f(c)\le y$. Suppose instead that $y<f(c)$. Put $\varepsilon=f(c)-y$, so $0<\varepsilon$, and take $\delta$ for this $\varepsilon$ as above. For $x\in[a,b]$ with $|c-x|<\delta$ we get $-(f(x)-f(c))\le|f(x)-f(c)|<\varepsilon$, and hence $y=f(c)-\varepsilon<f(x)$. Let $s\in S$; then $s\le c$, as $c$ is an upper bound of $S$. If $c-\delta<s$, then $0\le c-s<\delta$, so $|c-s|\le c-s<\delta$, and since $s\in[a,b]$ this gives $y<f(s)$, contradicting $f(s)\le y$. Hence $s\le c-\delta$, the order of $\mathbb{R}$ being total. So $c-\delta$ is an upper bound of $S$, whence $c\le c-\delta$ and $\delta\le0$, contradicting $0<\delta$. Therefore $f(c)\le y$, again by totality.

Step 2: $y\le f(c)$. Suppose instead that $f(c)<y$. Then $c\neq b$, for otherwise $f(b)=f(c)<y\le f(b)$; with $c\le b$ this gives $c<b$. Put $\varepsilon=y-f(c)$, so $0<\varepsilon$, and take $\delta$ for this $\varepsilon$ as above. Put $h=\delta/(1+1)$; by Rules of Arithmetic and Order in an Ordered Field (ix), applied to $0<\delta$, $0<h<\delta$. Choose $x_{0}$ as follows: if $c+h\le b$, let $x_{0}=c+h$, so that $x_{0}-c=h$; otherwise $b<c+h$, and let $x_{0}=b$, so that $x_{0}-c<h$. In both cases $c<x_{0}\le b$ and $0<x_{0}-c<\delta$; hence $a\le c<x_{0}$ gives $x_{0}\in[a,b]$, and $|c-x_{0}|=|-(x_{0}-c)|=|x_{0}-c|\le x_{0}-c<\delta$. So $f(x_{0})-f(c)\le|f(x_{0})-f(c)|<\varepsilon=y-f(c)$, that is $f(x_{0})<y$. Thus $x_{0}\in S$, and $x_{0}\le c$ as $c$ is an upper bound of $S$, contradicting $c<x_{0}$. Therefore $y\le f(c)$.

Steps 1 and 2 give $f(c)=y$, which proves Case 1.

Case 2: $f(b)\le y\le f(a)$. Let $k:[a,b]\to\mathbb{R}$ be the map $x\mapsto(-1)\cdot f(x)$ (Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (v)), so $k(x)=-f(x)$ by Rules of Arithmetic and Order in an Ordered Field (ii). Apply Continuity of Sums and Products of Real-Valued Functions on a Metric Space (v) with $(X,d)=(\mathbb{R},d_{\mathbb{R}})$, $A=[a,b]$, the point $x=a$, both of its maps $f$ and $g$ taken to be our $f$ (continuous at $a$ relative to $[a,b]$, being continuous on $[a,b]$), and its constant $c$ equal to $-1$; its map $cf$ is then $k$, so $k$ is continuous on $[a,b]$. By Rules of Arithmetic and Order in an Ordered Field (v), $k(a)=-f(a)\le-y\le-f(b)=k(b)$. Case 1, applied to $k$ and $-y$, gives $c\in[a,b]$ with $-f(c)=k(c)=-y$, and hence $f(c)=y$.

Proof of (ii). The image $f(I)$ is a subset of $\mathbb{R}$, and for every $v$, $v\in f(I)$ if and only if $v=f(x)$ for some $x\in I$, by Sets and Maps: Ordinary Notation (vii), The Image and the Preimage of a Class under a Class (i) and Values, Restriction and Equality of a Map, and the Map $u\mapsto t(u)$ as a Defined Set Symbol (i). By Intervals of the Real Line (i) we must show: if $u,w\in f(I)$ and $v\in\mathbb{R}$ satisfy $u\le v\le w$, then $v\in f(I)$. Let $x,z\in I$ with $u=f(x)$ and $w=f(z)$.

If $x=z$, then $u=w$, so $v=u$ and $v\in f(I)$. Otherwise let $s$ be the lesser and $t$ the greater of $x$ and $z$, so $s<t$ and $s,t\in I$. Every $r\in[s,t]$ satisfies $s\le r\le t$, hence $r\in I$ by Intervals of the Real Line (i); so $[s,t]\subseteq I$. Let $g=f|_{[s,t]}$ be the restriction of $f$ to $[s,t]$; it is a map $g:[s,t]\to\mathbb{R}$ with $g(r)=f(r)$ for every $r\in[s,t]$, by Domains and Restrictions of Maps (ii) and Domains and Restrictions of Maps (iii).

As $f$ is continuous on $I$, Restricting a Continuous Map to a Smaller Subset Preserves Continuity (ii), applied with $(X,d_X)=(Y,d_Y)=(\mathbb{R},d_{\mathbb{R}})$, $A=I$, $B=[s,t]$ and our $f$, shows that $g$ is continuous on $[s,t]$.

If $x<z$, then $s=x$, $t=z$, and $g(s)=u\le v\le w=g(t)$; if $z<x$, then $s=z$, $t=x$, and $g(t)=u\le v\le w=g(s)$. In either case (i), applied to $s<t$, $g$ and $v$, gives $c\in[s,t]$ with $g(c)=v$. Then $c\in I$ and $f(c)=g(c)=v$, so $v\in f(I)$. Hence $f(I)$ is an interval.