Understanding the idea of the proof of a question.

Question

Here is the question:

Proving some properties of the identity operator.

And here is the solution there:

Let $U=\{f: \|f\|_{\infty} <1\}$. This is an open set in $(C[0,1],\|.\|_{\infty}\})$. Its image is not open: $0$ is in the image; if the image is open then there exists $r>0$ such that $\|f-0\|_1 <r$ implies $f \in U$. Choose $n$ such that $\frac 1 n <r$ and define $f(x)=1-nx$ for $x \leq \frac 1 n$, $0$ for $x >\frac 1 n$. I will leave it to you to verify that $\|f\|_1 <r$ but $\|f\|_{\infty} =1$ (so $f \notin U$).

The solution is trying to prove that the identity operator is not open.

I know that the definition of an open mapping is:

A mapping $f: X \rightarrow Y$ from the topological space $X$ to the topological space $Y$ is said to be open provided the image of every open set in X is open in the topological space $f(X).$

My questions are:

1-So, I believe the solution is trying to contradict the definition of an open mapping i.e. it is trying to show that there is an open set in $(C[0,1], \|.\|_{\infty})$ whose image i.e. $I(U) = U$ is not open in $(C[0,1], \|.\|_1)$ ..... am I correct?

2- What is the image of $U$ is it $U$ itself but with $\|.\|_{1}$ instead?

2-Why is $U$ open set in $(C[0,1], \|.\|_{\infty})$? Is this because its definition is basically the definition of $B(0,1)$ and because every open ball is an open set ?

3- Is $0$ is in the image because the image is $U$ itself (but with $\|.\|_1)$ instead) and then what ? how do I prove that $0$ in the image?

4- Why is this the meaning of the image being open "if the image is open then there exists $r>0$ such that $\|f-0\|_1 <r$ implies $f \in U$"?

5- Now proving that $ $\|f\|_1

Here is my trial: $\|f\|_1 = \int_{0}^{1/n} (1-nx) dx = \int_{0}^{1/n}dx - \int_{0}^{1/n}(nx)dx = \frac{1}{n} - n \frac{1}{2n^2} = \frac{1}{2n} < r$ ... is my proof correct?

6- Lastly, it is not clear for me the idea of the function $f$ in proving that the image of $U$ is not open in $(C[0,1],\|.\|_{1}\}),$ could anyone explain it for me, please?

Answer 1

1. Yes
2. Yes. (2b) Also Yes, $U$ is just the unit ball for the $\infty$-norm and $I(U)$ contains the same functions as $U$, just equipped with a different norm.
3. Yes. By 2b, we just need to know that $0 \in U$ but clearly $\|0\|_\infty = 0 < 1$ so that $0 \in U$.
4. This isn't equivalent to the image being open but if the image is open then the statement in this point is true which is all you need. If $O$ is open and $x \in O$ then there is an $\varepsilon > 0$ such that $B(x, \varepsilon) \subseteq O$ by definition. Then apply that with $O = I(U)$ and $x = 0$.
5. Yes, this is correct.
6. In point 5, for arbitrary $r > 0$ you exhibit a function $f$ such that $\|f\|_1<r$ with $f \not \in U$. Point 4 tells you that if $I(U)$ is open then there is an $r > 0$ such that $\|f\|_1 < r$ implies $f \in U = I(U)$. This means that $I(U)$ cannot be open or these two statements would be contradictory.