# Compact sets and homeomorphisms

1. Apr 28, 2005

### Gunni

Hi there.

I'm taking a course in analysis and I was thinking about the relation between compact sets and homeomorphism. We know that if f is an onto and one-to-one homeomorphism then it follows that for every subset K:

K is compact in M <=> f(K) is compact in N

Now, does this go the other way too? That is, given that for every subset K that if K is compact then f(K) is compact, and vice versa, does it follow that f is an homeomorpism?

I've been trying to prove this by contradiction by assuming that f isn't continuous, taking a convergent series in K that isn't convergent in f(K) and trying to show somehow that this leads to that f(K) isn't compact, but so far no luck.

Any ideas on how to proceed, or if this in fact true?

2. Apr 28, 2005

### matt grime

What does homeomorphism mean? That f is continuous and that there is a continuous inverse, g, say.

So all we're doing is using the more basic fact that the continuous image of a compact set is compact.

Ie K compact implies f(K) compact, and f(K) compact imples gf(K)=K is compact.

3. Apr 28, 2005

### HallsofIvy

Staff Emeritus
Yes, matt, that's true but it's not what Gunni is asking.

His question is "Suppose f is a one-to-one function from M to N (having inverse g from N to M) such that f maps compact sets to compact sets and g maps compact sets to compact sets. Must f and g be continuous?"

4. Apr 28, 2005

### matt grime

Ah, sorry. My bad. In fact it's been a terrible day for maths full stop.

Let A be any algebraic variety with the zariski topology. Every function from A to A sends compact sets to compact sets, since every set is compact, yet not all bijections are continuous. (Note to self, say that but I've not proved it...)

Since the OP wanted to look at sequences, I think that we may be alright to restrict attention to metric spaces, when the answer is true, since in a metric space compact implies closed so f and g would send closed sets to closed sets, and hence open sets to open sets, thus they'd be continuous.

Last edited: Apr 28, 2005
5. Apr 28, 2005

### Gunni

Hmm... that's strange. I thought I managed to prove this earlier. Here's how:

Let (M,d) and (N,r) be metric spaces, and f:M -> N a one-to-one and onto function. Assume that for every subset K of M holds

K compact in M <=> f(K) compact in N

Let's show that f is continuous.
Take a convergent series (x_n) in M such that x_n -> x and let

$$K := \{ x_n : n\in \mathbb{N} \} \cup \{x\}$$

then K is compact and

$$f(K) = \{f(x_n) : n\in \mathbb{N} \} \cup \{f(x)\}$$

is also compact. Now let's take a subsequence $$f(x_{n_k})$$ in N and it's corresponding sequence in M, $$(x_{n_k})$$.

Now, $$f(x_{n_k})$$ has a Cauchy subsequence $$(f_{n_{k_i}})$$ which is convergent to f(y) (f(K) is compact). But the correspoding subsequence $$(x_{n_{k_i}})$$ is a subsequence of a convergent sequence, and so it converges to x. Therefore y=x, and

$$(f_{n_{k_i}}) \to f(x)$$

As this applies for every subsequence $$(f_{n_{k})$$, we get that

$$f(x_n) \to x$$

for every convergent sequence (x_n) in M. Therefore, f is continuous.

The steps for showing that f^-1 is continuous are exacly the same, so we conclude that f is a homeomorphism from M to N. QED.

I can't find an error in the argument above, but if you see one you're welcome to point it out to me.

6. Apr 28, 2005

### Gunni

By the way, I first tried arguing that since f sends compact sets into compact sets, f also sends closed sets into closed ones. This doesn't hold because it says nothing about a closed set which isn't totally bounded and all bets are off on that one.

Unless are no closed sets which aren't totally bounded, aside from the metric space itself?

And we are working in a metric space, I should have said that. Sorry.

7. Apr 29, 2005

### matt grime

Well, there you go metric spaces are easy, and I think I gave an open mapping proof earlier.

8. Apr 29, 2005

### Gunni

Does the open mapping argument work because of the reasons I posted above?

9. Apr 29, 2005

### matt grime

Actually my proof doesn't work, as you noted, and I'm not sure that yours does either.

Given x_n tending to x, then the image set f(x_n) union f(x) is compact, so sequentially compact so it has a subsequence that converges, but we don't know that f(x_n) itself is a convergent sequence, though by applying g to it you might be able to show that it must converge.

Last edited: Apr 29, 2005
10. Apr 29, 2005

### Hurkyl

Staff Emeritus
Any subset of a finite topological space is obviously compact. So, all you need to do to produce a counterexample is to find a map between finite topological spaces that isn't a homeomorphism.

How about the map f:{0, 1} --> {0} that sends everything to zero?

11. Apr 29, 2005

### Hurkyl

Staff Emeritus
Oh, here's another one.

Let X be the nonnegative reals, and let f simply be the inclusion X --> R.

Then, for any subset K of X, K is compact in X iff f(X) is compact in R.

12. Apr 29, 2005

### matt grime

But isn't f supposed to be invertible as a set map?

13. Apr 29, 2005

### HallsofIvy

Staff Emeritus
All compact sets are closed but not all closed sets are compact! How does "send compact sets to compact sets" show "send closed sets to closed sets"?

14. Apr 29, 2005

### matt grime

Yes, and that is where my proof is completely wrong (for metric spaces), as Gunni has pointed out, and I agreed previously.

The counter example in the non-metric space holds, though.
Another counter example there is, in the same vein:

Let R^2 be given two topologies: the usual zariski topology, and the product zariski topology. Then the natural identification of the underlying sets sends compact sets to compact sets but is not a homeomorphism.

Last edited: Apr 29, 2005
15. Apr 29, 2005

### Hurkyl

Staff Emeritus
Not the way the problem is stated...

16. Apr 29, 2005

### mansi

f being a homeomorphism...implies that f is invertible....

17. Apr 29, 2005

### Hurkyl

Staff Emeritus
The question was stated:

"That is, given that for every subset K that if K is compact then f(K) is compact, and vice versa, does it follow that f is an homeomorpism?"

The goal was to prove f is a homeomorphism, not to assume it. :tongue2:

Anyways, the finite topological space is easily rectifiable to be a bijection -- simply pick any two nonhomeomorphic spaces of the same finite cardinality. Say... {0, 1} with the discrete and indiscrete topologies.

18. Apr 29, 2005

### mansi

yeah...you're right...

19. Apr 30, 2005

### matt grime

But the question as you state it Hurkyl is trivially false. If we don't allow the extra hypothesis that f is invertible as a set map, then pretty much an inclusion will suffice.

The original question talked bout one toone and onto homeomorphisms, as if there were homeos that weren't bijections, and it implicityl assumed it was a metric space. I think we ought to assume the question requires f to at least be a bijection, otherwise, as I say, it is vacuously false.

Last edited: Apr 30, 2005
20. Apr 30, 2005

### Hurkyl

Staff Emeritus
That's why I revised my counterexample to be a set bijection.

Well, there's another way to revise the question:

Given a function f:X-->Y, is:
For any subset S of X, S is compact iff f(S) is compact.
For any subset T of Y, T is compact iff f-1(T) is compact.
equivalent to
Then f is a homeomorphism.

I'm having difficulty coming up with a counterexample when X and Y are infinite.