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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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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. –  user7530 Feb 26 '13 at 3:48

1 Answer 1

Show that linear functions are continuous and function composition of continuous functions is continuous.

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 just composing a linear function with a continuous function.

Edit: I should say I chose this approach because it is particularly easy to deal with from the topological point of view because it's quite easy to show preimages of open sets are open sets with linear functions, and with function composition of continuous functions.

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