How would one show that the set consisting of the monomials $1,x,x^2,…$ is a closed subset of the metric space $C[a,b]$ under the metric $d(a,b) = ||a-b|| =sup_{[0,1]}|a-b|$ ?
I considered its complement and had a hard time getting started; since given any element which is a sum of these terms (possibly with different coefficients, I don't see where to start in showing that an $\epsilon$-Ball of that element would be contained in the complement.
Proving every convergent sequence in my set of monomials has a limit in there is an option, but I really need some guidance on how to set up that kind of proof since I don't feel clever enough yet to pull of a proof with sequences.
Best Answer
Let $P=\{p_n\mid n\geqslant0\}$ where $p_n:[0,1]\to\mathbb R$, $x\mapsto x^n$. Assume that there exists a sequence $(f_n)_n\subseteq P$ and a function $f$ in $C[0,1]$ such that $f_n\to f$ uniformly. Thus $f_n=p_{\varphi(n)}$ for some function $\varphi:\mathbb N\to\mathbb N$. Either (i) there exists some $k$ such that $f_n=p_k$ for infinitely many $n$, or (ii) $\varphi(n)\to\infty$.
In case (i), a subsequence of $(f_n)_n$ is identically equal to $p_k$ hence it converges to $p_k$, and the whole sequence converges to $f$, hence $f=p_k$.
In case (ii), $\varphi(n)\to\infty$ hence $p_{\varphi(n)}(x)\to0$ if $x\ne1$ and $p_{\varphi(n)}(1)=1$ for every $n$, that is, $f_n\to g$ pointwise, where $g(x)=0$ for every $x\ne1$ and $g(1)=1$. Since $f_n\to f$ uniformly, $f_n\to f$ pointwise hence $f=g$. But $g$ is not in $C[0,1]$, this is absurd.
Finally, every sequence in $P$ which converges to a limit point in $C[0,1]$ has its limit in $P$, hence $P$ is closed in $C[0,1]$.