Real Analysis Problem involving Image of a Bounded Set

[brickcitie]

Homework Statement

Ok so I'm given that we have some function from R to R, that is continuous on all of R. I am asked if it is possible to find some BOUNDED subset of R such that the image of the set is not bounded. The professor gave the hint: look at closures.

Homework Equations

None.

The Attempt at a Solution

Been thinking about it for an hour. My first idea was to use 1/x, and let the subset of R be (0,1), but then I realized that 1/x is not continuous on all of R. I cannot think of any counter-example here, but I know that there is no theorem that states that if we have a continuous function the image of a bounded set is bounded, so I'm guessing there must be some counter-example to find.

Any help or direction would be wonderful. I thought about the closure hint and it got me nowhere at all.

 

[micromass]

Is the closure of a bounded set bounded? If yes, then the closure is closed and bounded...

 

[brickcitie]

I'm not quite sure how that helps though. How can I use that information to find a bounded set whose image is not bounded?

 

[micromass]

Well, maybe the image of a bounded set is always bounded...

 

[brickcitie]

Ok so then in that case I've been spinning my wheels trying to come up with a counter-example.

Any tips on how to begin the proof?

 

[micromass]

Well, start by showing that the closure of a bounded set is always closed and bounded.

Does "closed and bounded" not ring a bell?

 

[brickcitie]

Ok the closure of a bounded set is closed and bounded, and since I'm dealing with R then it is compact. I know that when dealing with a continuous function the image of a compact set is always compact. But that doesn't help if the set is open and bounded. This proof basically boils down to proving that using a continuous function the image of an open and bounded set must be bounded, in which case the image of any bounded set is bounded.

Not too sure how the closure comes into play here because you can just take the image of the set itself and not the closure. I appreciate all the responses, but I'm not sure if I understand where you are trying to steer me.

 

[micromass]

Not sure if you are aware of this, but a set which is not closed is not necessairily open. I say this, because I think you implicitly mentioned it.

As for the proof: Take a set A who is bounded. You have now shown that the closure of A is compact. We know that

A\subseteq \overline{A}

Can you show now that

f(A)\subseteq f(\overline{A})?

 

[brickcitie]

AHA! I do know that a set that is not closed is not necessarily open, yet I seem to always get carried away during proofs and forget that.

I understand where you were leading me now. A very clever proof. Thanks so much for your time and patience micro, you help is duly appreciated.