Which of the following sets are dense in $C[0,1]$ Which of the following sets are dense in $C[0,1]$ with respect to sup-norm topology?
$1$. {$f$$\in$ $C[0,1]$ : $f$ is a polynomial }
$2$. {$f$$\in$ $C[0,1]$ :$f(0)$=$0$}
$3$. {$f$$\in$ $C[0,1]$ :$f(0)$$\neq$$0$}
$4$. {$f$$\in$$C[0,1]$ :$\int_{0}^{1}f(x)dx$$=$5}
I am thinking to apply Stone-weierstrass theorem but I don't know how to use it properly. Please help!
 A: (1) follows directly from the stone-weirstrass theorem, since $[0,1]$ is a closed interval.
(2) is not dense, show that $f(x)=1$ is not a limit point
(3) Here, the closure of this set is $C[0,1]$, try showing that any $g\in C[0,1]$ with $g(0)=0$ is a limit point.
(4) Not dense, again show that $f(x)=1$ is not a limit point.
A: Hint for 2 :
Let, $S=\{f\in C[0,1]:f(0)\not =0\}$.
Then $S=\{f\in C[0,1]:f(0)<0\}\cup \{f\in C[0,1]:f(0)>0\}$.
So, $\overline S=\{f\in C[0,1]:f(0)\le0\}\cup \{f\in C[0,1]:f(0)\ge0\}=C[0,1]$.
So, $S$ is dense in $C[0,1]$.
Again as $S$ is the union of two open sets so it is open subset of $C[0,1]$.
Hence its complement is nowhere dense.
That is, the set $\{f\in C[0,1]:f(0)=0\}$ is nowhere dense.
Similarly option 4 is nowhere dense.
A: Answer: $1,\, 3$
Proof:


*

*Stone-Weierstrass theorem states explicitly  the dense property of the polynomials in $(C[0,1],\parallel\cdot\parallel_\infty)$ 

*It is not dense. Because it is a proper closed subset of  $(C[0,1],\parallel\cdot\parallel_\infty)$ as it is the inverse image of $\{0\}$ under the continuous map $f\mapsto f(0)$.

*It is dense. It is obvious that every element of that set is a limit point of that set. So now, let $f(0)=0$. Then the sequence of functions $f_n(x)=f(x)+\frac{1}{n}$ converges to $f$ with respect to supnorm and $f_n(0)=\frac{1}{n}\neq 0$.

*It is not dense. Because it is   the inverse of the set $\{5\}$ under the continuous map $f\mapsto \int_0^1 f(x)dx$. And hence it is a proper closed set.
A: Hint : 


*

*Stone-weierstrass theorem

*Think about $\{x,x^2,x^3\cdots\}$ 

*Think about $\{a+x^i : a\in \mathbb{R}\backslash \{0\}; i\in \mathbb{N}\}$

*I have no hint for this as of now but i believe this is too small to generate $C[0,1]$

