Is the preimage of a bounded set also bounded?

Question

I need to prove the following statement:

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ a continuous function and $B \subseteq \mathbb{C}$ bounded, implies, that the set $A=f^{-1}(B)$ to be bounded.

I do know that the statement is true for a continuous function on closed sets.

Is it possible, that this statement is also true for just bounded sets; and if not, what is the argumentation?

score 6 · Accepted Answer · answered Dec 19 '14 at 16:44

6

Consider constant function $f(z)=c$ for all $z$, then $\{c\}$ is bounded, but $f^{-1}(\{c\})=\mathbb{C}$.

answered Dec 19 '14 at 16:44

agha

10,038

So in case my set $B$ is closed the statement is true. – nando Dec 19 '14 at 16:48
Just to get it right:
A bounded image does not imply, that the preimage must be bounded.
– nando Dec 19 '14 at 16:56
What about the fact, that a constant function has no inverse function? – nando Dec 19 '14 at 17:00
Yes, a bounded image doesn't imply that preimage is bounded. But what do you mean with inverse function? Do you want to know that if $f$ has an inverse function your statement is true? – agha Dec 19 '14 at 17:08
Yes, that is what I am thinking about. – nando Dec 19 '14 at 17:10

hmakholm left over Monica · Answer 2 · 2014-12-19T17:55:33.110

3

For a nonconstant (in fact injective) counterexample, since you're not requiring $f$ to be differentiable, you can just do something like $$ f(z) = \frac{z}{1+|z|} $$ which is a continuous function on a closed set (namely, all of $\mathbb C$). The preimage of the open unit disc is then all of $\mathbb C$.

edited Dec 19 '14 at 17:55

answered Dec 19 '14 at 17:07

hmakholm left over Monica

286,031

Thanks a lot, I think get closer now! – nando Dec 19 '14 at 17:09
1

@PhoemueX: That is still not injective: You get $f(\frac12)=f(2)=\frac 25$. On the other hand $\frac{x}{1+|x|}$ would work, if the tanh offends you. – hmakholm left over Monica Dec 19 '14 at 17:27
@HenningMakholm: Ok, maybe I should stop thinking about math for tonight. Thank you very much! – PhoemueX Dec 19 '14 at 17:34

Is the preimage of a bounded set also bounded?

2 Answers2