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) $.

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.

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?

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.