Quotient space of $C[0,1]$ Construct the quotient space of $C[0,1]$ with the subspace
$$
M:=\left\{f \in C[0,1]: f\left(\frac{1}{2} \pm \frac{1}{2^n}\right)=0, n\in \Bbb N\right\}.
$$
I really don't know how to solve it, I would appreciate a hint or example to help me understand it.
 A: You have a sequence $\{x_n\}$ and $M=\{f\in C[0,1]:\ f(x_n)=0,\ n\in\mathbb N\}$.  So two functions will be equal in the quotient if they agree on all $x_n$. Thus a class will be defined by its values in the set  $\{x_n\}$. Being bounded, it looks like we can identify the quotient with $\ell^\infty(\mathbb N)$. But there is an added factor, which is that $\{x_n\}$ as given in the question has an accumulation point, $t=1/2$. So the values $f(x_n)$ converge to $f(1/2)$ since $f$ is continuous, and then the candidate for the quotient is $c$, the space of convergent sequences.
Now, let's do it formally. Define $\pi:C[0,1]/M\to c$ by $\pi(f+M)=\{f(x_n)\}_n$.

*

*Well defined: if $f-g\in M$, then $f(x_n)=g(x_n)$ for all $n$. And, as $x_n\to 1/2$ and $f$ is continuous, $f(x_n)\to f(1/2)$, so $\pi(f+M)$ is convergent.


*Linearity is obvious, as $\pi$ is an evaluation.


*Injective: if $f(x_n)=0$ for all $n$, then $f\in M$


*Surjective: given $y\in c$, we can construct $f$ as linear segments joining the points $(x_n,y_n)$. As $x_n\to1/2$, we define $f(1/2)=\lim_nx_n$.
