If $f:[a,b] \to \mathbb{R}$ continous function then $f([a,b])$ is bounded (duplicate and verifying attempt)
I am trying to prove this problem, can you check if this is true?
Here is my attempt:
Let $a_1=a$, $b_1=b$ and $I_1=[a_1,b_1]$, then let $a_2=a_1$, $b_2=\frac{b_1+ a_1}{2}$ and $I_2=[a_2, b_2]$ and let $a_{n+1}=a_n$, $b_{n+1}=\frac{b_n+ a_n}{2}$, recursively. Lastly, $I_n=[a_n,b_n]\subseteq [a,b]$. We can obserwe $a_n \to a$ and $b_n \to a$ . If we write the $\epsilon-\delta$ definition of the continuity at a, then$\forall \epsilon >0$; $\exists \delta >0$ such that $|x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon$. Since $I_n$ intervals are neighbors of $a$, we can modify the definition as follows $|f(x)-f(a)|<\epsilon$ ($\forall n>0$ and $\forall x \in I_n$) also we can get $|f(x)|<1+|f(a)|$ if we take $\epsilon=1$. Now assuming $f$ is not bounded on $[a,b]$, $n_0>0$ and $\exists x \in I_{n_0}$ $|f(x_0)|>n_0$ . In the definition we gave using $I_n$ intervals which are neighbors of $a$, it continues to work if we take $n=n_0>|f(a)|+1$ but here we get $|f(x_0)|>n_0>|f(a)|+1$ this means $|f(x_0)|>|f(a)|+1$ and $|f(x_0)-f(a)|>1$ This gives the contradiction that the function f is not continuous for $\epsilon=1 $ at point $a$. Then our assumption that f is not bounded is false, which proves the correctness of our claim that $f([a,b])$ is bounded.
I am not sure after the last part $n=n_0>|f(a)|+1$
This is the point your solution goes off the rails. You haven't used continuity at any point other than $a$, but at this point, only using $f$ is continuous at $a$, you've already concluded that $f$ is bounded.
The strongest correct statement I can give in place of that one is
which effectively is just saying $f$ is bounded on some $I_n$.
There are some later statements that need doctored too, in order to make any sense of them, but it doesn't actually matter because there is no valid way to conclude $f([a,b])$ is bounded if you only use the assumption that $f$ is continuous at $a$ and not on the entirety of $[a,b]$.