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Here's my latest try:

Suppose it's not. It means there are two non-empty disjoint open sets $A,B$ such that $A \cup B = C[0,1] \setminus P$. "Take" $f \in A, g \in B$ such that $f(x)<g(x) (\forall x \in [0,1])$. Call $f_1$ some non-polynomial function such that $f(x)<f_1(x)<g(x)(\forall x \in [0,1])$. If $f_1 \in A$, call $f_2$ some non-polynomial function such that $f_1(x)<f_2(x)<g(x)(\forall x \in [0,1])$. Otherwise, call $f_2$ some non-polynomial function such that $f(x)<f_2(x)<f_1(x)(\forall x \in [0,1])$. Repeat the process and construct a sequence $(f_n)$ in $C[0,1] \setminus P$ (that is convergent in $C[0,1]$) which limit is both on $\overline A$ and $\overline B$, a contradiction...if the sequence is also convergent on $C[0,1] \setminus P$!

Can I guarantee that this limit is not a polynomial? Since $C[0,1] \setminus P$ is not complete, I find it hard to complete my proof. Thanks.

Hilario Fernandes
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  • 'not connected' does not mean you have nonempty disjoint open (in the ambient space) sets which make up the set you are looking at. 'Open' refers to the relative topology. As an example remove $(0,1)$ from $[0,1]$. What you get is not connected. Also, if you find sets like the ones you have described, how do you know that you can find $f<g$ as claimed? – Thomas Dec 10 '15 at 18:28
  • -I'm not sure I understood what you said. The relative topology of the subspace is not the relative topology of the space where the sets are replaced by their intersections with the subspace? I mean, the open sets of a subspace aren't the open sets of the space intersected with the subspace? – Hilario Fernandes Dec 10 '15 at 18:44
  • -If you suppose that a function $f(x)$ and some translation ($f(x)+k, k \in \mathbb{R}$) aren't on the same open set, you can use the argument of the question to construct a sequence which limit is in $\overline A \cap \overline B$. Hence the prop. would be proved. It means you have to consider the case where ${f(x)+k \mid k \in \mathbb{R}} \subset A$ (or $B$). In this case, you can translate functions freely and guarantee that $f<g$ and $f \in A, g \in B$. – Hilario Fernandes Dec 10 '15 at 18:45

1 Answers1

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Here is a tedious construction that shows a path between two non polynomial functions. The idea is very simple, the details tiresome.

Suppose $f$ is not a polynomial. Then there is some $a \in (0,1)$ such that the restriction $f \mid_{[a,1]}$ is not a polynomial. To see this, suppose $f \mid_{[a,1]}$ is a polynomial for all $a \in (0,1)$, then, since a polynomial is uniquely defined by its values on a non empty interval, we see that $f \mid_{[a,1]}$ is the same polynomial regardless of $a$, and, by continuity, we see that $f$ is a polynomial which is a contradiction.

Now suppose$ f_1,f_2$ are not polynomials, then we from the previous remarks it follows that there are $a_1,a_2$ such that the respective restrictions are not polynomials. Let $a=\min(a_1,a_2)$, then $f_1 \mid_{[a,1]}$, $f_2 \mid_{[a,1]}$ are not polynomials.

Let $\phi_t = f_1-t f_1(a)$. We see that $\phi_t$ is not a polynomial for all $t$, hence there is a path from $f_1$ to $f_1'=f_1-f_1(a)$ that doesn't pass through any polynomial. We note that $f_1'(a) = 0$.

The same procedure shows that there is a path from $f_2$ to $f_2'=f_2-f_2(0)$ that doesn't pass through any polynomial and $f_2'(a) = 0$.

Some notation: Suppose $q:[0,a] \to \mathbb{R}$, $w:[a,1] \to \mathbb{R}$ are such that $q(a)=w(a) = 0$, then let $(q,w)$ be the function $(q,w): [0,1] \to \mathbb{R}$ such that $(q,w)(x) = \begin{cases} q(x), & x \in [0,a] \\ w(x), & x \in [a,1] \end{cases}$.

Now let $g:[0,a] \to \mathbb{R}$ be any function that is not a polynomial (for example, $g(x) = \sin ( {t \over a} \pi)$). I will use the same symbols $f_1',f_2'$ to represent the relevant restrictions below.

Then the collection of functions $t \mapsto (f_1'+t(g-f_1')), f_1')$ shows a path from $f_1'$ to $(g,f_1')$ that does not pass through any polynomial (since $f_1' \mid_{[a,1]}$ is not a polynomial).

Then the collection of functions $t \mapsto (g, f_1'+t(f_2'-f_1'))$ shows a path from $(g,f_1')$ to $(g,f_2 ')$ that does not pass through any polynomial (since $g$ is not a polynomial).

Finally, the collection of functions $t \mapsto (g+t(f_2'-g),f_2')$ shows a path from $(g,f_2 ')$ to $f_2'$ that does not pass through any polynomial (since $f_2' \mid_{[a,1]}$ is not a polynomial).

copper.hat
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