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Prove that all polynomials from $\mathbb{R}$ to $\mathbb{R}$ are continuous. Now this is from a topological point of view.

I thought that maybe induction would work here?

Initial Case: $f(x) = a_{0}$ where $a_{0} \in \mathbb{N}$. This is continuous.

Inductive Case: $f(x) = a_{0} + a_{1}x + a_{2}x^2 + ... + a_{n}x^n + ...$

Since the sum of continuous functions is continuous, this implies every term is continuous. Makes sense. Feel like theirs a lot of holes in my reasoning though. For one I am not sure how to show the constant $a_{0}$ is continuous from a topological view.

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  • $\begingroup$ The constant function is particularly easy: the inverse image of any open set $S$ is either the empty set or $\mathbb{R}$, depending on whether $a_0\in S$. Both of these are open. $\endgroup$
    – user7530
    Feb 26, 2013 at 3:48

1 Answer 1

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There are a few ways you can go about this, one approach might be from first principles prove that if $f(x)$ is continuous, then so is $xf(x) + c$.

Then consider the following iterations:

$$ p_0(x) = a_n $$ $$ p_1(x) = x p_0(x) + a_{n-1} $$ $$ p_2(x) = x p_1(x) + a_{n-2} $$ $$ \dots $$ $$ p_n(x) = f(x) $$

At each step you're multiplying a continuous function by $x$ and then adding a constant and so an inductive argument can go from there.

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    $\begingroup$ Your claim that each step is the composition of a linear function with a continuous function is incorrect. $\endgroup$
    – MPW
    Dec 24, 2020 at 4:04
  • $\begingroup$ Good point, i'll edit the answer $\endgroup$
    – muzzlator
    Jun 30, 2021 at 9:30

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