(I'm sorry to post in one thread, but I didn't want to cluster up the forums with multiple threads)

The following are 8 possible outcomes of continuous mappings and its image and preimages. I have several questions regarding notations as well.

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Let $\displaystyle A \subseteq \mathbb{R}^n $ and $\displaystyle F:A \rightarrow \mathbb{R} $ be continuous.

Images:

1.If A is closed, is F(A) closed?

False. Let $\displaystyle A = \mathbb{R} $ and define $\displaystyle f(x) = tan^{-1} x $. A is closed in R but $\displaystyle f(A) = \left( \frac{- \pi}{2}, \frac{\pi}{2} \right) $ is open and not closed.

2.If A is open, is F(A) open?

I believe this is true but I am not sure.

3.If A is compact, is F(A) compact?

True. Let $\displaystyle u_{k} $ be a sequence in F(A). Let $\displaystyle u_{k} = F(v_{k}) $. Since A is sequentially compact, then there is a subsequence $\displaystyle v_{k_j} \rightarrow v \in A$. Thus, $\displaystyle {u_{k_j}} = {F(v_{k_j})}$

4.If A is bounded, is F(A) bounded?

False. Let $\displaystyle A = \left( \frac{- \pi}{2}, \frac{\pi}{2} \right)$ and $\displaystyle f(x) = tanx$. Since $\displaystyle A = \left( \frac{- \pi}{2}, \frac{\pi}{2} \right)$ is bounded, but $\displaystyle f:tan \left( \frac{- \pi}{2}, \frac{\pi}{2} \right) \rightarrow R $ is unbounded.

**If suppose it was $\displaystyle f:R^n \rightarrow R^m $ is bounded, then would $\displaystyle f(A)= \mathbb{R}^m$ be bounded?

Preimages:

5.If A is closed, is$\displaystyle F^{-1}(A) $closed?

I believe this is true. Since $\displaystyle F^{-1}(A^{c}) = (f^{-1}(A))^c $. So, if A is closed, then $\displaystyle A^{c}$ is open. Iffis continuous, then the preimage of every closed set in R is closed in A. By definition.

6.If A is open, is$\displaystyle F^{-1}(A) $open?

True. Iffis continuous, then the preimage of every open set in R is open in A by definition.

7.If A is compact, is$\displaystyle F^{-1}(A) $compact?

False, let A = [0, 1] which is compact because it is closed and bounded. Then let f(x) = 1/2, the the image $\displaystyle F(A) = \mathbb{R} $ which is not compact.

8.If A is bounded, is$\displaystyle F^{-1}(A) $bounded?

I do not know. I believe this is false, but I can not think of a counterexample.

Regarding notations, what happens if instead of $\displaystyle f:A \rightarrow \mathbb{R}$ is continuous, it was $\displaystyle f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ is continuous. Would the results change?

Thank you for reading. Any help is greatly appreciated.