Proof of Continuity of Identity Maps, Compositions and Reciprocals

Proof of (i). Let $\iota:A\to X$ be the map $u\mapsto u$, so that $\iota(u)=u$ for $u\in A$ (Sets and Maps: Ordinary Notation (v)). Let $x\in A$ and let $\varepsilon>0$ be real; put $\delta=\varepsilon$. Every $y\in A$ with $d(x,y)<\delta$ satisfies $d(\iota(y),\iota(x))=d(y,x)=d(x,y)<\varepsilon$, by Metric Space (iv). So $\iota$ is continuous at $x$ relative to $A$ by Continuous Map Between Metric Spaces (i); as $x\in A$ was arbitrary, $\iota$ is continuous on $A$ by Continuous Map Between Metric Spaces (ii).

Proof of (ii). Let $f$ be continuous at $x\in A$ relative to $A$ and $g$ continuous at $f(x)$ relative to $B$, and let $\varepsilon>0$ be real. The choices are made in this order. First, Continuous Map Between Metric Spaces (i) for $g$ at $f(x)$ gives a real $\eta>0$ such that every $s\in B$ with $d_{Y}(f(x),s)<\eta$ satisfies $d_{Z}(g(s),g(f(x)))<\varepsilon$. Second, the same clause for $f$ at $x$, applied to $\eta$, gives a real $\delta>0$ such that every $u\in A$ with $d_{X}(x,u)<\delta$ satisfies $d_{Y}(f(u),f(x))<\eta$. Let $u\in A$ with $d_{X}(x,u)<\delta$. Then $s=f(u)$ lies in $B$ and $d_{Y}(f(x),s)=d_{Y}(f(u),f(x))<\eta$, by Metric Space (iv); so

$$d_{Z}\bigl((g\circ f)(u),(g\circ f)(x)\bigr)=d_{Z}\bigl(g(f(u)),g(f(x))\bigr)<\varepsilon .$$

Hence $g\circ f$ is continuous at $x$ relative to $A$, by Continuous Map Between Metric Spaces (i). If $f$ is continuous on $A$ and $g$ on $B$, then for every $x\in A$, $f$ is continuous at $x$ relative to $A$ and $g$ at $f(x)\in B$ relative to $B$, by Continuous Map Between Metric Spaces (ii); so $g\circ f$ is continuous at every $x\in A$ relative to $A$, that is, continuous on $A$.

Proof of (iii). Put $r=1/g$. Continuity relative to $E$ is used in the form of Continuity of Real Functions on Subsets of ℝ in ε-δ Form (i). For $t\in E$, $g(t)\neq0$, so $r(t)=1/g(t)=g(t)^{-1}$ and $g(t)\,r(t)=1$ (Difference and Quotient Identities in ℝ (ii)), and $|r(t)|=|g(t)|^{-1}$ (Difference and Quotient Identities in ℝ (iii)). For $t\in E$,

$$(\dagger)\qquad r(t)-r(x)=\bigl(g(x)-g(t)\bigr)\,r(t)\,r(x),$$

since the right side is $g(x)\,r(x)\,r(t)-g(t)\,r(t)\,r(x)=r(t)-r(x)$, by Rules of Arithmetic in a Commutative Ring: Zero, Signs and Squares, and No Zero Divisors in a Field (ii) for $(u-v)\,w=u\,w-v\,w$ and by $g(x)\,r(x)=1=g(t)\,r(t)$.

Let $g$ be continuous at $x$ relative to $E$. Put $a=|g(x)|$; then $0<a$, as $0\le a$ and $a\neq0$ (Rules of Arithmetic and Order in an Ordered Field (x)), and $0<a/2$ and $a/2+a/2=a$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (vii). Put $K=(a/2)^{-1}\,a^{-1}$; then $0<K$ by Rules of Arithmetic and Order in an Ordered Field (viii) and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (iv). Let $\varepsilon>0$ be real; then $0<\varepsilon\,K^{-1}$ by the same two clauses. The choices are made in this order. First, continuity of $g$ at $x$, applied to $a/2$, gives a real $\delta_{1}>0$ such that every $t\in E$ with $|t-x|<\delta_{1}$ satisfies $|g(t)-g(x)|<a/2$. Second, the same continuity, applied to $\varepsilon\,K^{-1}$, gives a real $\delta_{2}>0$ such that every $t\in E$ with $|t-x|<\delta_{2}$ satisfies $|g(t)-g(x)|<\varepsilon\,K^{-1}$. Third, take a real $\delta$ with $0<\delta$, $\delta\le\delta_{1}$ and $\delta\le\delta_{2}$, such as $\min\{\delta_{1},\delta_{2}\}$.

Let $t\in E$ with $|t-x|<\delta$, and put $z=|g(t)-g(x)|$; then $|t-x|<\delta_{1}$ and $|t-x|<\delta_{2}$, by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (i). Also $|g(x)-g(t)|=z$, as $g(x)-g(t)=-(g(t)-g(x))$ (Difference and Quotient Identities in ℝ (i)) and $|-w|=|w|$ (Rules of Arithmetic and Order in an Ordered Field (x)). If $z=0$, then $g(t)-g(x)=0$ by Rules of Arithmetic and Order in an Ordered Field (x), so $g(t)=g(x)$, $r(t)=r(x)$ and $|r(t)-r(x)|=|0|=0<\varepsilon$. Let now $z\neq0$, so that $0<z$. By Rules of Arithmetic and Order in an Ordered Field (xi) and the first choice,

$$a=\bigl|\bigl(g(x)-g(t)\bigr)+g(t)\bigr|\le z+|g(t)|<a/2+|g(t)|,$$

the strict inequality by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (ii); so $a/2+a/2=a<a/2+|g(t)|$ by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (i), and $a/2<|g(t)|$ by Rules of Arithmetic and Order in an Ordered Field (iv). Hence $|r(t)|=|g(t)|^{-1}<(a/2)^{-1}$ by Rules of Arithmetic and Order in an Ordered Field (viii). By $(\dagger)$, Rules of Arithmetic and Order in an Ordered Field (x) and $|r(x)|=a^{-1}$,

$$|r(t)-r(x)|=z\,|r(t)|\,a^{-1}<z\,(a/2)^{-1}\,a^{-1}=z\,K<\varepsilon\,K^{-1}\,K=\varepsilon ,$$

where the first inequality multiplies $|r(t)|<(a/2)^{-1}$ by $z\,a^{-1}$, positive by Rules of Arithmetic and Order in an Ordered Field (viii) and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (iv), and the second multiplies $z<\varepsilon\,K^{-1}$, the second choice, by $K>0$, both by Rules of Arithmetic and Order in an Ordered Field (vi). So $r$ is continuous at $x$ relative to $E$, by Continuity of Real Functions on Subsets of ℝ in ε-δ Form (i).

If $g$ is continuous on $E$, then it is continuous at every $x\in E$ relative to $E$ (Continuous Map Between Metric Spaces (ii)); by the above, so is $r$, which is therefore continuous on $E$, by the same clause.