Proof of Continuity of the Identity Map, of Powers, and of Polynomial Functions on a Subset of the Real Line

Throughout, a map $g:E\to\mathbb{R}$ is continuous at $x\in E$ if and only if for every $\varepsilon>0$ there is $\delta>0$ such that every $y\in E$ with $|y-x|<\delta$ satisfies $|g(y)-g(x)|<\varepsilon$; this is Continuous Map Between Metric Spaces written out through the metric $d_{\mathbb{R}}$.

1. (Identity.) Let $x\in E$ and let $\varepsilon>0$; take $\delta=\varepsilon$. Every $y\in E$ with $|y-x|<\delta$ satisfies $|y-x|<\varepsilon$. Hence the identity map is continuous at $x$ relative to $E$, and since $x\in E$ was arbitrary, it is continuous on $E$.

2. (Powers.) Write $\pi_n:E\to\mathbb{R}$ for the map $x\mapsto x^{n}$, and let

$$T=\{n\in\mathbb{N}:\pi_n\text{ is continuous on }E\} .$$

By clause 1 of Properties of Natural Number Powers in a Field, $x^{1}=x$ for every $x\in\mathbb{R}$, so $\pi_1$ is the identity map of $E$ and $1\in T$ by claim 1.

Suppose $m\in T$. By the same clause, $x^{S(m)}=x^{m}\,x$ for every $x$, so $\pi_{S(m)}$ is the pointwise product of $\pi_m$ and the identity map. Both are continuous on $E$, so by clauses 3 and 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space the product $\pi_{S(m)}$ is continuous on $E$; that is, $S(m)\in T$.

By the principle of induction, $T=\mathbb{N}$.

3. (Polynomial functions.) Let $p$ be a polynomial function on $\mathbb{R}$ and let $(N,c_0,c)$ be a system of coefficients for it as in Polynomial Function on a Field, so that $c:[N]\to\mathbb{R}$ is defined on the initial segment determined by $N$ and

$$p(x)=c_0+\sum_{k=1}^{N}c_k\,x^{k}\qquad(x\in\mathbb{R}) .$$

For $j\in[N]$ define $\Sigma_j:E\to\mathbb{R}$ by $\Sigma_j(x)=\sum_{k=1}^{j}c_k x^{k}$, the finite sum of the first $j$ terms. Let

$$U=\{j\in\mathbb{N}: j\notin[N]\ \text{ or }\ \Sigma_j\text{ is continuous on }E\} .$$

We show $U=\mathbb{N}$ by induction. By clause 1 of Basic Properties of Initial Segments of the Natural Numbers we have $1\in[N]$, and the recursion in Finite Sum Notation in a Field gives $\Sigma_1(x)=c_1x^{1}$, which is a constant multiple of $\pi_1$ and hence continuous on $E$ by claim 2 together with clauses 4 and 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space; so $1\in U$.

Suppose $j\in U$. If $S(j)\notin[N]$ then $S(j)\in U$ by definition. Otherwise $S(j)\in[N]$, that is, $S(j)\le N$ by Initial Segment of the Natural Numbers. By clause 5 of Properties of the Order on the Natural Numbers, $j<S(j)$, so $j\le S(j)\le N$ and hence $j\le N$ by clause 1 of that lemma; therefore $j\in[N]$, and $\Sigma_j$ is continuous on $E$ because $j\in U$. By the recursion in Finite Sum Notation in a Field,

$$\Sigma_{S(j)}(x)=\Sigma_j(x)+c_{S(j)}\,x^{S(j)}\qquad(x\in E) .$$

The second summand is a constant multiple of $\pi_{S(j)}$ and is continuous on $E$ by claim 2 with clauses 4 and 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space. So $\Sigma_{S(j)}$ is a pointwise sum of two maps continuous on $E$, hence continuous on $E$ by clauses 2 and 5 of that theorem, and $S(j)\in U$.

By the principle of induction $U=\mathbb{N}$. Since $N\in[N]$, the map $\Sigma_N$ is continuous on $E$.

Finally, $p|_{E}$ is the pointwise sum of the constant map on $E$ with value $c_0$, continuous by clause 1 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, and of $\Sigma_N$. By clauses 2 and 5 of that theorem, $p|_{E}$ is continuous on $E$.