Proof of Continuity of the Identity Map, of Powers, and of Polynomial Functions on a Subset of the Real Line
lemmalem:continuity-identity-polynomial-real-2026aThe identity map is continuous with delta equal to epsilon; powers follow by induction on the exponent using the recursion defining a finite product, and polynomial functions by induction along the initial segment of coefficients using the recursion defining a finite sum.
Throughout, a map is continuous at if and only if for every there is such that every with satisfies ; this is Continuous Map Between Metric Spaces written out through the metric .
1. (Identity.) Let and let ; take . Every with satisfies . Hence the identity map is continuous at relative to , and since was arbitrary, it is continuous on .
2. (Powers.) Write for the map , and let
By clause 1 of Properties of Natural Number Powers in a Field, for every , so is the identity map of and by claim 1.
Suppose . By the same clause, for every , so is the pointwise product of and the identity map. Both are continuous on , so by clauses 3 and 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space the product is continuous on ; that is, .
By the principle of induction, .
3. (Polynomial functions.) Let be a polynomial function on and let be a system of coefficients for it as in Polynomial Function on a Field, so that is defined on the initial segment determined by and
For define by , the finite sum of the first terms. Let
We show by induction. By clause 1 of Basic Properties of Initial Segments of the Natural Numbers we have , and the recursion in Finite Sum Notation in a Field gives , which is a constant multiple of and hence continuous on by claim 2 together with clauses 4 and 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space; so .
Suppose . If then by definition. Otherwise , that is, by Initial Segment of the Natural Numbers. By clause 5 of Properties of the Order on the Natural Numbers, , so and hence by clause 1 of that lemma; therefore , and is continuous on because . By the recursion in Finite Sum Notation in a Field,
The second summand is a constant multiple of and is continuous on by claim 2 with clauses 4 and 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space. So is a pointwise sum of two maps continuous on , hence continuous on by clauses 2 and 5 of that theorem, and .
By the principle of induction . Since , the map is continuous on .
Finally, is the pointwise sum of the constant map on with value , continuous by clause 1 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, and of . By clauses 2 and 5 of that theorem, is continuous on .
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Prerequisites
19b16af6-4276-4634-a9c8-ac49e77d63b6