Which of the following sets are dense in $C[0,1]$

Answer: $1,\, 3$

Proof:

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

2. 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 $f\mapsto f(0)$.

3. 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 $f_n(0)=\frac{1}{n}\neq 0$.

4. 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.

(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.

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.