Is $C[0,1] \setminus P$ (where $P$ is the set of polynomials) connected? 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.
 A: 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).
