I would guess that polynomials are dense in space of continuous functions? (This wasn't covered in book or notes however.)
Yes, they are. This is the Stone-Weierstrass theorem. But if you haven't covered that yet, then don't worry about it - you can handle it using Taylor series.
I hadn't thought about Taylor series being a polynomial.
Each partial sum of a Taylor series has only finitely many terms, and is therefore a polynomial! And the sum of the series is the limit of the sequence of partial sums.
So Taylor series that converges to exp^t would be an example to use.
Taylor Expansion of exp^t is 1+t+...+t^n/n!+...
I could use p(t)=1+t+...+t^n/n!.
Right, although I would call it p_n(t) since you're building a sequence.
And guess that it converges to p=Taylor expansion.
\lim_{n \rightarrow \infty} p_n(t) = p(t) = e^t
By the way, you might want to call the limit something other than p(t) since the notation suggests that it's a polynomial, and the whole point is that you don't want it to be a polynomial. But that's just a nitpick.
Need to show that inf[p(t)-p]-->0.
Don't you mean sup, not inf? And writing it more carefully, it would be
\sup_{t \in [0,1]} [p_n(t) - p(t)] \rightarrow 0
as n \rightarrow \infty
i.e. you need to establish that the convergence is uniform. Do you know any theorems about the uniformity of convergence of Taylor series (or, more generally, power series)?
Then since p is not an element of P[0,1], the space is not complete.
Correct.
Question: where do Cauchy sequences come into this proof?
Good question. They haven't appeared explicitly yet, but having established that p_n \rightarrow p uniformly in the larger space of continuous functions, you immediately know that the sequence p_n is uniformly Cauchy. (If you haven't covered this fact already, you should prove it, but it's an easy proof.)
