Problem regarding convergency: $\sum_{k=0}^{\infty}\frac{x^k}{k!} \to e^x$

Question

Question. Let $X$ be the space of all polynomials in one variable,with real coefficients. If $p=a_0+a_1x+\dots+a_nx^n\in X$, define $$|p|=|a_0|+|a_1|+\dots+|a_n|,$$ which gives the metric $d(p,q)=|p-q|$ on $X$.

Does the metric space $X$ is complete?

Now, this question has been answered 'false' here.

According to that answer let $p_n(x)=\sum_{k=0}^{n}\frac{x^k}{k!}$ ...

I know this is due to the fact of Taylor series (theorem) $$\sum_{k=0}^{\infty} \frac{x^k}{k!}= e^x ,\ \text{*as*} \ n \longrightarrow \infty \ ,\forall \ x \in \mathbb{R}: ~p_n \to e^x$$ But still I have some questions:

1. When we say $p_n \to e^x$ as $n \to \infty$.then In which space this convergency is happening...?? I mean w.r.to which norm? Does the space here is $X=C(\mathbb{R}):= \text{Space of continuous functions on}~ \mathbb{R}$ ...But then which norm?? I mean "sup-norm" shouldn't work here as $e^x$ is not bounded...!!

2. Now how does this convergency (i.e. $p_n\to e^x$) remains independent of norm put on the space?

Please help me to clarify these confusions. Correct me if I'm wrong. Thank you.

Answer 1

The statement that $p_n$ converges to $e^{x}$ and hence it does not converge in the given space is just suppose to be a hint and not a complete proof. No norm is used in saying that $p_n$ converges to $e^{x}$. You have to give a rigorous proof along the following lines: if a sequence of polynomials converges in the given norm then the coefficients converge. So if $p_n$ converges to a polynomial $p$ then its coefficients have to be $\frac 1 {k!}$, $k=0,1,2,...$. In particular no coefficient can be $0$ which contradicts the fact that $p$ is a polynomial. To be more explicit suppose $p$ has degree $m$. Since each each $p_n$ with $n >m$ has $\frac 1 {(m+1)!}$ as the $m+1$-st coefficient the same must be true of $p$ but $p$ does not have this term.

Answer 2

Proving that a metric space is not complete means showing that there exists a Cauchy sequence (https://en.wikipedia.org/wiki/Cauchy_sequence) of its elements which does not have a limit in this space. So basically, you first want to prove that the elements of your sequence become arbitrary close (that for $ \forall \epsilon > 0, \exists N: \forall n, m > N, |p_n - p_m| < \epsilon $). Also, since your metric is defined only on polynomials, you can not directly prove that $ p_n $ converges to $ e^x $ (because ($e^x - p_n$) is not a polynomial). The easiest way around is to suppose that the given sequence has some polynomial as its limit, and come to a contradiction. (as it is done in the above comment)