Proof of The Square of a Nonnegative Continuous Real-Valued Function is Continuous

We write $s<t$ for real numbers to mean $s\le t$ and $s\ne t$, $s-t$ for $s+(-t)$, and $|\cdot|$ for the absolute value, whose properties we take from Properties of the Absolute Value in an Ordered Field; order arithmetic is taken from Elementary Arithmetic in an Ordered Field and Elementary Order Arithmetic in an Ordered Field, the identity $(p-q)(p+q)=p\,p-q\,q$ is claim 4 of Zero Products and Elementary Identities in a Field and the identity $(-p)q=-(pq)$ is claim 2 of that lemma, while $(t\,r^{-1})r=t$ for $r\ne 0$ is elementary arithmetic in the underlying field. By The Absolute Value Metric on the Real Line, $d_{\mathbb{R}}(p,q)=|p-q|$.

Fix $x\in A$ and a real number $\varepsilon$ with $0<\varepsilon$. Set $2=1+1$ and $s=1+2f(x)$.

Step 1: the constants. By claim 8 of Elementary Order Arithmetic in an Ordered Field we have $0<2$, so $0\le 2$; since $0\le f(x)$, claim 5 of Elementary Arithmetic in an Ordered Field gives $0=2\cdot 0\le 2f(x)$. Claim 6 of Elementary Order Arithmetic in an Ordered Field gives $0<1$, so claim 3 of that lemma gives $0=0+0<1+2f(x)=s$. By claim 7 the inverse $s^{-1}$ exists and $0<s^{-1}$, and by claim 5 we get $0<\varepsilon\,s^{-1}$. By claim 9 of the same lemma there is a real number $\eta$ with $\eta\le 1$, $\eta\le\varepsilon\,s^{-1}$, and $\eta$ equal to $1$ or to $\varepsilon\,s^{-1}$; in either case $0<\eta$.

Step 2: choice of $\delta$. Since $f$ is continuous at $x$ relative to $A$, there is a real number $\delta$ with $0<\delta$ such that every $y\in A$ with $d(x,y)<\delta$ satisfies $|f(y)-f(x)|<\eta$.

Step 3: the estimate. Fix such a $y$ and put $a=f(y)-f(x)$ and $b=f(y)+f(x)$, so that $ab=f^{2}(y)-f^{2}(x)$.

By claim 2 of Elementary Arithmetic in an Ordered Field we have $0\le b$, since $0\le f(y)$ and $0\le f(x)$. Moreover $b=a+2f(x)$, and claim 3 of Properties of the Absolute Value in an Ordered Field gives $a\le|a|$, while $|a|<\eta\le 1$; claim 2 of Elementary Order Arithmetic in an Ordered Field then gives $a<1$, and claim 1 of that lemma gives

$$b=a+2f(x)<1+2f(x)=s ,$$

so in particular $b\le s$.

Now, using claim 5 of Elementary Arithmetic in an Ordered Field twice (first with $a\le|a|$ and $0\le b$, then with $b\le s$ and $0\le|a|$, the latter by claim 1 of Properties of the Absolute Value in an Ordered Field), claim 10 of Elementary Order Arithmetic in an Ordered Field with $|a|<\eta$ and $0<s$, and claim 5 of Elementary Arithmetic in an Ordered Field once more with $\eta\le\varepsilon\,s^{-1}$ and $0\le s$,

$$ab\le|a|\,b\le|a|\,s<\eta\,s\le(\varepsilon\,s^{-1})s=\varepsilon .$$

By transitivity and claim 2 of Elementary Order Arithmetic in an Ordered Field this gives $ab<\varepsilon$.

The same chain applies with $a$ replaced by $-a$: claims 2 and 3 of Properties of the Absolute Value in an Ordered Field give $-a\le|-a|=|a|$, so $(-a)b\le|a|\,b$ and therefore $(-a)b<\varepsilon$. Since $(-a)b=-(ab)$, claim 4 of Elementary Order Arithmetic in an Ordered Field turns this into $-\varepsilon<ab$.

Step 4: conclusion. From $-\varepsilon<ab$ and $ab<\varepsilon$, claim 9 of Properties of the Absolute Value in an Ordered Field gives $|ab|<\varepsilon$, that is

$$d_{\mathbb{R}}\bigl(f^{2}(y),f^{2}(x)\bigr)=\bigl|f^{2}(y)-f^{2}(x)\bigr|<\varepsilon .$$

Hence $f^{2}$ is continuous at $x$ relative to $A$, and since $x\in A$ was arbitrary, $f^{2}$ is continuous on $A$ relative to $A$.