# Prove that the metric space is incomplete?

Question: Let $$P[0,1]$$ be the set of all polynomials defined on $$[0,1]$$. A metric is defined $$P[0,1]$$ by $$d(P_1, P_2)=\sup\limits_{0\leq x\leq 1}|P_1(x)-P_2(x)|$$. Then show that this metric space is incomplete.

We know that a metric space is complete if every Cauchy sequence in the metric space is convergent in that metric space. I cannot understand what will be the approach? Here every element in $$P[0,1]$$ is polynomial. How can I get sequence from $$P[0,1]$$.

• Consider $e^x$ on $[0,1]$. Why is it not a polynomial? Does it have a series expansion? Can you find polynomials going to $e^x$ uniformly? – астон вілла олоф мэллбэрг Feb 9 at 6:06

Take any real analytic function (function which equals its power series), but which is not a polynomial. For instance $$\ln (1+x),e^x,\sin x,\cos x$$ etc.

Then it can be approximated by polynomials.

• My mistake. Thanks. – Chris Custer Feb 9 at 6:59

So, what happens if we try to find the limit that should be there? Well, from the definition of that distance, $$\left|P_1(x)-P_2(x)\right| \le d(P_1,P_2)$$ for any fixed $$x$$. As such, if a sequence $$Q_n$$ is Cauchy, its values $$Q_n(x)$$ at a particular point are also a Cauchy sequence - in $$\mathbb{R}$$. That sequence has a limit, because $$\mathbb{R}$$ is complete.

So then, we have a candidate limit: given a Cauchy sequence $$Q_n$$, the function $$Q(x)=\lim_{n\to\infty} Q_n(x)$$ "should" be the limit of the $$Q_n$$. What could go wrong? Two possibilities:

• That's a pointwise limit, not directly using the metric. What if it's not convergent in the metric sense - that $$d(Q_n,Q)\neq 0$$?
• What if $$Q$$ isn't a polynomial? We defined it as a function, after all.

As it turns out, the first issue isn't a problem - we indeed have $$d(Q_n,Q)\to 0$$. The second, though? There are lots of ways to get a sequence of polynomials that converges to something that's not a polynomial.

For an example, how about a geometric series? Consider the sequence $$Q_n(x) = 1+\frac{x}{2}+\frac{x^2}{4}+\cdots+\frac{x^n}{2^n} = \sum_{k=0}^n \left(\frac x2\right)^n = \frac{1-\frac{x^{n+1}}{2^{n+1}}}{1-\frac x2}$$ Since $$\frac{x^{n+1}}{2^{n+1}} \le \frac1{2^{n+1}}\to 0$$, we have $$Q_n(x)\to Q(x)=\frac{1}{1-\frac x2}=\frac{2}{2-x}$$, with error $$\sup_x |Q_n(x)-Q(x)| \le 2^{-n}$$ and $$d(Q_n,Q_m)\le 2^{-\min(m,n)}$$. That's a Cauchy sequence that doesn't converge in $$P[0,1]$$, because its limit isn't a polynomial.

As a side note, there's a theorem: for any continuous function $$f$$ on the closed interval, there is a sequence of polynomials $$f_n$$ that converges uniformly (that is, $$\lim_n d(f_n,f) = 0$$ in the metric here) to $$f$$. We can get basically everything.