Hey!!

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be continuous and $A,B\subseteq \mathbb{R}$. Prove or disprove the following:

- $A$ closed $\Rightarrow$ $f(A)$ closed
- $B$ closed $\Rightarrow$ $f^{-1}(B)$ closed
- $A$ bounded $\Rightarrow$ $f(A)$ bounded
- $B$ bounded $\Rightarrow$ $f^{-1}(B)$ bounded

I have done the following:

We have that $D$ is closed $\iff$ for each convergent sequence $(x_n)$ in $D$ it holds that $\displaystyle{\lim_{n\rightarrow \infty} x_n\in D}$.

We also have that $f: D\rightarrow \mathbb{R}$ is bounded $\iff$ $f(D)$ is bounded. This is equivalent to $\exists c\geq 0 \ \ \forall x\in D : |f(x)|\leq c$.

- What can we do in this case?

$$$$- Let $(x_n)$ be a convergent sequence in $f^{-1}(B)$. We want to show that $\displaystyle{\lim_{n\rightarrow \infty}x_n\in f^{-1}(B)}$.

Since $f$ is continuous, we have that $\displaystyle{f\left (\lim_{n\rightarrow \infty} x_n\right )=\lim_{n\rightarrow \infty} f(x_n)}$.

Since $(x_n)\in f^{-1}(B)$ and $f$ is continuous we have that $f(x_n)\in B$. Is this correct?

Since $B$ is closed we have that $\displaystyle{\lim_{n\rightarrow \infty}f(x_n)\in B}$.

Therefore, we have that $\displaystyle{f\left (\lim_{n\rightarrow \infty}x_n\right )\in B}$. From that it implies that $\displaystyle{\lim_{n\rightarrow \infty}x_n\in f^{-1}(B)}$. This means that $f^{-1}(B)$ is closed.

Is this correct?

$$$$- Since $A$ is bounded, we have that it is upper and under bounded. There are $x,y$ such that $$x\leq a \leq y , \ \ \text{ for all } a\in A$$ Since $f$ is continuous we have that $$f(x)\leq f(a) \leq f(y) , \ \ \forall a\in A$$

This means that in $A$ the function $f$ is bounded.

Therefore, $f(A)$ is bounded.

Is this correct?

$$$$- What can we do in this case?