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

**jsndacruz** · April 15th 2012, 06:21 PM

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 $f^-1 (C)$ .

**jsndacruz** · April 16th 2012, 08:13 AM

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.

**Plato** · April 16th 2012, 08:56 AM

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.

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?

**jsndacruz** · April 16th 2012, 09:11 AM

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.

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

LinkBack

Thread Tools

Display

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

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

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

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

Similar Math Help Forum Discussions

Image and Inverse Images for Projection Functions

Images of sets

Open and closed sets and functions

polynomial functions question (images incl)

Images of Compact and Connected Sets

Search Tags