# Homework Help: Inverse Image of a Compact Set - Bounded?

1. Nov 18, 2007

### varygoode

[SOLVED] Inverse Image of a Compact Set -- Bounded?

Problem:

Let f : X → Y be a continuous function, K ⊂ Y - compact set. Is it true that f$$^{-1}$$(K)– the inverse image of a compact set– is bounded? Prove or provide counterexample.

Questions Generated:

1. Why does compactness matter? (I know it does.)
2. Does knowing that the inverse of a continuous function sends closed sets to closed sets (and open to open) matter?

Solution Ideas:

Well, I just feel like this is false and has a counterexample. But I'm not sure. I was thinking that somehow I can show the inverse sends compact sets to compact sets, and that compact sets are closed and bounded. But this is only true in $$\mathbb{R}$$$$^{n}$$. So I'm not really sure.

2. Nov 18, 2007

### morphism

Firstly I should point out that a continuous map need NOT pull back compact sets to compact sets; maps that do this are called proper.

Now as to the question at hand, try to think about how nice continuous functions on general metric spaces are. How nice they are depends on the metric, no? So try to give X some metric that doesn't restrict the set of continuous maps on X, like the discrete metric.

3. Nov 19, 2007

### varygoode

If I choose X as the discrete metric, won't any function sending any set to X then be discrete, as X is?

Because then it would be bounded, because all sets in the discrete metric are bounded by 1.

So I'm not exactly sure what the counterexample you were aiming at there is.

4. Nov 20, 2007

### morphism

Oops - you're right of course. I had misinterpreted the question!

But anyway, there are lots of counterexamples - take for instance any constant function from R to R (in the standard metric).

5. Nov 20, 2007

### HallsofIvy

What happens if you use a constant function? (specific example deleted- surely you can think of your own!)

6. Nov 20, 2007

### varygoode

Ah, this is good stuff. Thanks folks.

So I can choose some $$f:\mathbb{R} \rightarrow \mathbb{R}$$ such that $$\{f(x)=0 \mid x\epsilon\mathbb{R}\}$$, for instance. Then $$f$$ is compact easily since an open cover for it is $$\mathbb{R}$$ and a finite subcover is obvious. Then any $$f^{-1}$$ would send this $$\{0\}$$ back to $$\mathbb{R}$$, which is unbounded.

Thanks again! This is solved.

Last edited: Nov 20, 2007
7. Nov 20, 2007

### matt grime

That is odd notation. f is a function, not a subset of R. It is not compact. The image of f is a compact subset of R. However, the definition of compact is that *any* open cover has a finite subcover. You only wrote down one cover, that contained one set, not an arbitrary open cover. But a one point set is trivially compact, anyway, just not for the reasons you set out.

8. Nov 20, 2007

### varygoode

Ah, yes, all true. However, a union of all open covers should be $$\mathbb{R}$$. And $$f$$ was a typo, should be "the image of $$f^{-1}$$".

9. Nov 21, 2007

### morphism

What does that have to do with anything?

To show that {0} is compact you have to take an arbitrary open cover of {0} and show that it admits a finite subcover. But this is easy: at least one of the sets in any open cover will contain {0}, and any single such set will be a sufficient finite subcover.

10. Nov 21, 2007

### HallsofIvy

Using f-1({0}) instead of f in your earlier response, that is simply R= f-1({0}) which is definitely NOT compact! In fact, you say that solves your problem because R is not bounded. Surely you know that every compact set, in a metric space, is bounded!

11. Nov 21, 2007

### varygoode

HallsofIvy, it is only the {0} that needs to be compact, not $$f^{-1}(\{0\})$$. So it's all good.

12. Nov 21, 2007

### HallsofIvy

Yes, in fact, to satisfy this problem, it can't be compact!

I was referring to your statement, "Then f is compact easily since an open cover for it is R and a finite subcover is obvious." and your subseuent statement, "f was a typo, should be "the image of f-1", which, since you were talking about {0}, I took to mean f-1({0})= R.

Are you clear on the definition of "compact"? The fact that {R} is an open cover for R and is already finite is irrelevant. In fact, {R} is a finite open cover for every set! Are you claiming that every set is compact? In order for a set, A, to be compact, it is not just necessary that there be SOME open cover that has a finite subcover. That's true for any set. What must be true is that every open cover of A contains a finite subcover.

Last edited by a moderator: Nov 21, 2007
13. Nov 21, 2007

### varygoode

I guess I was unclear on the definition, wow. Thanks for making me understand. Sorry about all that.