Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function

Justification of the identity in [L6]. Let $u,v\in \mathbb{R}$. By totality either $u\ge v$ or $v\ge u$. If $u\ge v$ then $u-v\ge 0$, so $\mathrm{\mid }u-v\mathrm{\mid }=u-v$ and $\frac{1}{2}(u+v+\mathrm{\mid }u-v\mathrm{\mid })=\frac{1}{2}(2u)=u=\max \{u,v\}$, while $\frac{1}{2}(u+v-\mathrm{\mid }u-v\mathrm{\mid })=v=\min \{u,v\}$. If $v\ge u$ the same computation with the roles exchanged applies, since $\mathrm{\mid }u-v\mathrm{\mid }=v-u$ there.

The isolated case. Suppose $c$ is an isolated point of $A$, say $N_{\eta }(c)\cap A=\{c\}$ with $\eta >0$ real. Then every function on $A$ is continuous at $c$ by [L2], which gives claims 1, 2 and 3 at once. For claim 4, assume $g(c)\mathrm{\ne }0$; then $c\in A_0$, and $N_{\eta }(c)\cap A_0\subseteq N_{\eta }(c)\cap A=\{c\}$ with $c$ in the left-hand side, so $c$ is an isolated point of $A_0$ and every function on $A_0$, in particular $(f\mathrm{/}g){\mathrm{\mid }}_{A_0}$, is continuous at $c$.

Claim 2, at any point of $A$. Let a real $\epsilon >0$ be given and let $\delta >0$ be as in [L1] for $f$ and this $\epsilon$. For $x\in A$ with $\mathrm{\mid }x-c\mathrm{\mid }<\delta$ the reverse triangle inequality gives $\mid \mathrm{\mid }f(x)\mathrm{\mid }-\mathrm{\mid }f(c)\mathrm{\mid }\mid \le \mathrm{\mid }f(x)-f(c)\mathrm{\mid }<\epsilon$. So $\mathrm{\mid }f\mathrm{\mid }$ is continuous at $c$, and no case distinction was needed.

Claim 5, constants and the identity. If $f$ is constant then $\mathrm{\mid }f(x)-f(c)\mathrm{\mid }=0<\epsilon$ for every $x\in A$ and every real $\epsilon >0$, so any $\delta >0$ serves. For the identity, given a real $\epsilon >0$ take $\delta :=\epsilon$: every $x\in A$ with $\mathrm{\mid }x-c\mathrm{\mid }<\delta$ has $\mathrm{\mid }\mathrm{i}\mathrm{d}(x)-\mathrm{i}\mathrm{d}(c)\mathrm{\mid }=\mathrm{\mid }x-c\mathrm{\mid }<\epsilon$. Both are continuous at every point of $A$.

The limit-point case, claim 1. Suppose $c$ is a limit point of $A$. By [L2] the limits of $f$ and of $g$ at $c$ exist and equal $f(c)$ and $g(c)$. By [L3] the limits of $f+g$, $\alpha f$ and $fg$ at $c$ exist and equal $f(c)+g(c)$, $\alpha f(c)$ and $f(c)g(c)$, which are exactly the values of those three functions at $c$; by [L2] again, each of them is continuous at $c$.

The limit-point case, claim 4. Suppose $c$ is a limit point of $A$ and $g(c)\mathrm{\ne }0$. Then $c\in A_0$, and by [L4] the point $c$ is a limit point of $A_0$. By [L3] the limit of $(f\mathrm{/}g){\mathrm{\mid }}_{A_0}$ at $c$ exists and equals $f(c)\mathrm{/}g(c)$, which is the value of $(f\mathrm{/}g){\mathrm{\mid }}_{A_0}$ at $c$; by [L2] applied on the domain $A_0$, that function is continuous at $c$.

Claims 1 and 4 in general. By [L2] the point $c$ is either isolated in $A$ or a limit point of $A$; step 1.2 settles the first case and steps 1.5 and 1.6 the second. So claims 1 and 4 hold as stated.

Claim 3. By claim 1 the function $f-g=f+(-1)g$ is continuous at $c$, by step 1.3 so is $\mathrm{\mid }f-g\mathrm{\mid }$, and by claim 1 again so are $f+g+\mathrm{\mid }f-g\mathrm{\mid }$ and its scalar multiple by $1\mathrm{/}2$. By step 1.1 that scalar multiple is the function $x\mapsto \max \{f(x),g(x)\}$, so the maximum is continuous at $c$; the same argument with $-\mathrm{\mid }f-g\mathrm{\mid }$ gives the minimum.

Claim 5, powers and polynomials. The map $x\mapsto x^0$ is the constant $1$ and $x\mapsto x^1$ is the identity, both continuous on $A$ by step 1.4; and if $x\mapsto x^n$ is continuous on $A$ then so is $x\mapsto x^{n+1}=x^n\cdot x$, being a product of two functions continuous on $A$ by step 2.1. By induction on $n$, $x\mapsto x^n$ is continuous on $A$ for every $n\in \mathbb{N}$. A polynomial function $a_0+a_{1}x+\cdots +a_n{x}^n$ is obtained from these by finitely many scalar multiplications and additions, each of which preserves continuity by step 2.1.

Claims 1 to 5 are proved, all of them at an arbitrary point $c$ of $A$ and therefore, applied at every point, on the whole of $A$; and no sequence and no choice principle was used.