The space of all finite-degree polynomials $\mathbb{P}$ is not complete in any norm.

Question

Let $\mathbb{P}$ denote the space of all finite degree polynomials in one variable. Show that $\mathbb{P}$ is never complete with respect to $P_1$ norm, i.e., $\|\cdot\|_1$ by giving an example Cauchy sequence that does not converge inside $\mathbb{P}$?

Answer 1

A correct solution:

The set $\{1,x,x^2,\cdots\}$ is a basis for $\mathbb{P}$. Consider the subspaces $X_n= \langle 1, x, \cdots, x^n \rangle$. Each $X_n$ is closed because every finite dimensional normed space is complete. Also, $X_n$ is a proper subset of $X$ and has empty interior because if it contains an open ball, the homogeneity of the norm allows it to cover all elements in $X$. Also, $X = \cup_{n=1}^{\infty} X_n$. Therefore, $X$ cannot be complete because otherwise we have just shown that it is a countable union of nowhere dense sets, which contradicts Baire's category theorem.

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The initial answer which is wrong (even though I loved it):

Given any norm $\|\cdot\|$ on $\mathbb{P}$, the statement is true. Since it is a norm, and the polynomial $B_k(x)=x^k$ is a non-zero polynomial for all $k \in \{0,1,\cdots\}$, the real number $\|x^k\|$ is not zero for any $k$. Now consider the sequence $$P_n(x) = \sum_{k=0}^n \frac{1}{\|x^k\|2^k}x^k$$

By the triangle inequality, we have that

$$\|P_n(x)-P_m(x)\| = \left\|\sum_{k=m}^n \frac{1}{\|x^k\|2^k}x^k \right\|\leq \sum_{k=m}^n\frac{1}{2^k\|x^k\|}\|x^k\|=\sum_{k=m}^n\frac{1}{2^k}$$

The last sum can be made arbitrarily small because $\sum_{k=0}^{\infty}\frac{1}{2^k}=2$. Hence, the sequence $\big(P_n(x)\big)_{n=1}^{\infty}$ is Cauchy but it turns out, as pointed out by Eric Wofsey, that whether it converges in $\mathbb{P}$ or not is not a simple matter.