# Continuous mapping - closed, open, compact, bounded

• Oct 12th 2008, 09:09 AM
Paperwings
Continuous mapping - closed, open, compact, bounded
(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. If f is 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. If f is 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.
• Oct 12th 2008, 10:22 AM
ThePerfectHacker
Quote:

Originally Posted by Paperwings
(
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.

This is false.

Quote:

2. If A is open, is F(A) open?
I believe this is true but I am not sure.
No, consider the constant function.

Quote:

3. If A is compact, is F(A) compact?
True.
Yes, this is true. This is a well-known result.

Quote:

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.
False again.

Quote:

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. If f is continuous, then the preimage of every closed set in R is closed in A. By definition.
With pre-images this is true.

Quote:

6. If A is open, is $\displaystyle F^{-1}(A)$ open?
True. If f is continuous, then the preimage of every open set in R is open in A by definition.
Again true.

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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.
This is false since the next one is false.

Quote:

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.
Consider $\displaystyle f:\mathbb{R}\to \mathbb{R}$ defined by $\displaystyle f(x) = 0$. Then $\displaystyle \{ 0 \}$ is bounded but its preimage is unbounded.