# Continuous functions: Images and Pre-images of closed and open sets

• Apr 15th 2012, 06:21 PM
jsndacruz
Continuous functions: Images and Pre-images of closed and open sets
I'm having a bit of difficulty seeing whether a function or its inverse preserves whether a set's image/pre-image remains open of closed. There are 5 parts for which I must either prove or give a counter-example. I cannot find an example for 3. and can't think of how to prove 4. I put all of them down though - just in case I messed anything up.

Setup:
Let $\displaystyle f:\mathbb{R} \to \mathbb{R}$ be continuous. Also, let $\displaystyle S \subseteq \mathbb{R}$ and $\displaystyle T \subseteq \mathbb{R}.$

1. If $\displaystyle S$ is open, then $\displaystyle f(S)$ is open.
False: If $\displaystyle f(x) = c$ for some constant $\displaystyle c$, then its range is closed no matter the pre-image.

2. If $\displaystyle S$ is closed, then $\displaystyle f(S)$ is closed.
False: Using $\displaystyle f(x)=e^x$ and $\displaystyle S = ( - {\infty} , 0 )$, then $\displaystyle f(S) = (0,1)$.

3. If $\displaystyle S$ is bounded, then $\displaystyle f(S)$ is bounded.
False: But I'm having difficulty thinking of an example.

4. If $\displaystyle T$ is open, then $\displaystyle f^-1 (T)$ is open.
True: But again, I'm not sure where to turn for the proof.

5. If $\displaystyle T$ is closed, then $\displaystyle f^-1 (T)$ is closed.
True: If I had the above, I'd use its complement to show this one holds, but since I don't... Let $\displaystyle x \in f^-1 (T)$ and let $\displaystyle x_n$ be a sequence of elements from $\displaystyle x \in f^-1 (T)$ that converges to $\displaystyle x$. Since $\displaystyle f$ is continuous, $\displaystyle f(x_n)$ converges to $\displaystyle f(x)$. Since $\displaystyle C$ is closed, $\displaystyle f(x)$ is in C. So, $\displaystyle x$ is in $\displaystyle f^-1 (C)$.
• Apr 16th 2012, 08:13 AM
jsndacruz
Re: Continuous functions: Images and Pre-images of closed and open sets
I realize I can use the complement of 5. to get 4. - I would, however, still appreciate a direct method if anyone knows.

As for 3., I might be going crazy .. intuitively, I think the statement is true. After all, since $\displaystyle f$ is continuous on $\displaystyle \mathbb{R}$, it is continuous on $\displaystyle S$. If it were an interval l could look at $\displaystyle [a,b]$, where $\displaystyle a=inf(S)$ and $\displaystyle b=sup(S)$, I could use continuity to show that the function is bounded on the interval. But $\displaystyle S$ is a set - and I'm not sure how that changes things.
• Apr 16th 2012, 08:56 AM
Plato
Re: Continuous functions: Images and Pre-images of closed and open sets
Quote:

Originally Posted by jsndacruz
Setup:
Let $\displaystyle f:\mathbb{R} \to \mathbb{R}$ be continuous. Also, let $\displaystyle S \subseteq \mathbb{R}$ and $\displaystyle T \subseteq \mathbb{R}.$

1. If $\displaystyle S$ is open, then $\displaystyle f(S)$ is open.
False: If $\displaystyle f(x) = c$ for some constant $\displaystyle c$, then its range is closed no matter the pre-image.

2. If $\displaystyle S$ is closed, then $\displaystyle f(S)$ is closed.
False: Using $\displaystyle f(x)=e^x$ and $\displaystyle S = ( - {\infty} , 0 )$, then $\displaystyle f(S) = (0,1)$.

3. If $\displaystyle S$ is bounded, then $\displaystyle f(S)$ is bounded.
False: But I'm having difficulty thinking of an example.

4. If $\displaystyle T$ is open, then $\displaystyle f^-1 (T)$ is open.
True: But again, I'm not sure where to turn for the proof.

5. If $\displaystyle T$ is closed, then $\displaystyle f^-1 (T)$ is closed.
True: If I had the above, I'd use its complement to show this one holds, but since I don't... Let $\displaystyle x \in f^-1 (T)$ and let $\displaystyle x_n$ be a sequence of elements from $\displaystyle x \in f^-1 (T)$ that converges to $\displaystyle x$. Since $\displaystyle f$ is continuous, $\displaystyle f(x_==nn)$ converges to $\displaystyle f(x)$. Since $\displaystyle C$ is closed, $\displaystyle f(x)$ is in C. So, $\displaystyle x$ is in $\displaystyle f^-1 (C)$.

Be careful in #2. $\displaystyle (-\infty,0)$ is not close. But $\displaystyle (-\infty,0]$ is closed, while $\displaystyle (0,1]$ is not.

Quote:

Originally Posted by jsndacruz
I realize I can use the complement of 5. to get 4. - I would, however, still appreciate a direct method if anyone knows.
As for 3., I might be going crazy .. intuitively, I think the statement is true. After all, since $\displaystyle f$ is continuous on $\displaystyle \mathbb{R}$, it is continuous on $\displaystyle S$. If it were an interval l could look at $\displaystyle [a,b]$, where $\displaystyle a=inf(S)$ and $\displaystyle b=sup(S)$, I could use continuity to show that the function is bounded on the interval. But $\displaystyle S$ is a set - and I'm not sure how that changes things.

Every bounded set of real numbers is the subset of a closed finite interval.
So it is the subset of a compact set. The function is continuous. Thus?
• Apr 16th 2012, 09:11 AM
jsndacruz
Re: Continuous functions: Images and Pre-images of closed and open sets
Ah - for #2 I meant to use square brackets. Thanks for paying attention to me!

My professor never taught us about 'compact' sets - my grasp on topology is really poor. But I think I know what you're getting at: since S is a subset of a closed interval, it is sufficient to show that the function is bounded on that interval, since the subset of a bounded set must be bounded.