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

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

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

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

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

5. If $T$ is closed, then $f^-1 (T)$ is closed.
True: If I had the above, I'd use its complement to show this one holds, but since I don't... Let $x \in f^-1 (T)$ and let $x_n$ be a sequence of elements from $x \in f^-1 (T)$ that converges to $x$. Since $f$ is continuous, $f(x_n)$ converges to $f(x)$. Since $C$ is closed, $f(x)$ is in C. So, $x$ is in $f^-1 (C)$.
• 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 $f$ is continuous on $\mathbb{R}$, it is continuous on $S$. If it were an interval l could look at $[a,b]$, where $a=inf(S)$ and $b=sup(S)$, I could use continuity to show that the function is bounded on the interval. But $S$ 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 $f:\mathbb{R} \to \mathbb{R}$ be continuous. Also, let $S \subseteq \mathbb{R}$ and $T \subseteq \mathbb{R}.$

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

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

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

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

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

Be careful in #2. $(-\infty,0)$ is not close. But $(-\infty,0]$ is closed, while $(0,1]$ 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 $f$ is continuous on $\mathbb{R}$, it is continuous on $S$. If it were an interval l could look at $[a,b]$, where $a=inf(S)$ and $b=sup(S)$, I could use continuity to show that the function is bounded on the interval. But $S$ 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.