Compact sets and homeomorphisms

[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?

 

[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.

 

[HallsofIvy]

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?"

 

[matt grime]

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

Anyway, the answer is no.

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.

 

[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.

 

[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.

 

[matt grime]

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

 

[Gunni]

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

 

[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.

 

[Hurkyl]

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?

 

[Hurkyl]

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.

 

[matt grime]

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

 

[HallsofIvy]

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

Anyway, the answer is no.

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.

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"?

 

[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.

 

[Hurkyl]

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

Not the way the problem is stated...

 

[mansi]

Not the way the problem is stated...

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

 

[Hurkyl]

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.

 

[mansi]

yeah...you're right...

 

[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.

 

[Hurkyl]

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.

 

[matt grime]

Well, there are a whole host of counter examples for non-metric spaces, but metric ones are being a pain in the bum. Partly becuase i) I don't think it's true in general, and ii) If we ask: "if there are two metrics on the same space which share the same compact sets, are they equivalant?" then this is true in almost every case you can think of: all norms on a finite dimensional vector space are equivalent, if compact is equivalent to closed, eg if the spaces are totally bounded, then its true. I think there may be something doable with "compact exhaustions" that limit counter examples too.

As I know very little about p-adic analysis, perhaps someone else can suggest if this is a way forward:

Take two different p-adic valuations on Q. Are these inequivalent topologies? I seem to remember they are, what are the compact sets? Are they the same in each topology? If so, this is a counter example.

Of course there may well be other ways to generate counter examples, or it could even be true. But that just seems unlikely to me.

 

[Gunni]

Well, I'm out of my league here. I'm only in my first undergrad year and I haven't taken any topology yet (although I have Munkres' book and intend to put it to good use in the summer), so I'll have to study some before a proof or counterexample along those lines will make sense to me.

This question comes from a large project from one of my courses. We split up into groups of three or four and were supposed to solve three big projects over the semester. One of those concerned homeomorphisms; in one part of it we were given five statements about homeomorphisms which we were able to prove were equivalent. Then in the next section two more followed, one about continuous functions and the one above.

We could prove that both followed from the fact that f is a one-to-one and onto homeomorphism, and that the one about continuous functions also led to f being a homeomorphism. That's where the question comes from, does this one lead back too?

Not one of us out of a twenty- or thirty something students has a counter-example or a convincing proof, so at least it's not obvious (I hope).

By the way, this isn't really a homework question since the final exam was today and we turned the assignments in last week. :)

 

[matt grime]

Again you're saying 1-1 and onto as if there were homeos that weren't bijections.

I'm still not sure, even from that, exactly what the question you're asking is. There have been several answers to different interpretations of your (ambiguous) question.

Would you mind stating clearly what your question is *exactly*

As Hurkyl (I think) said, if the question was simeply:

suppose f is a function between topological spaces satisfying K is compact if and only if f(K) is compact, does it follow that f is a homeomorphism?

Then the answer is trivially "no", eg the inclusion of [0,1] into R

 

[Gunni]

I'm sorry, I didn't realize my question was ambiguous.

But if there are no significant differences between metric and topological spaces, then your understanding is correct. Thank you for your answers. I'll try to be more specific in the future.

 

[matt grime]

There are huge differences between metric and topological spaces. The metric topology is just about the nicest there is.

It wasn't clear if you were requiring that f is an invertible map, you see. At least not to me. I was assuming that was one of the hypotheses, and that your question was:

"if f is a bijeciton between (metric) topological spaces such that f(K) is compact iff K is compact, is f a homeomorphism."

which turned out to be a tricky question to answer. it is false for non-metric topologies, and I think it is false for metric ones, though I can't find an example; your proof showed that given a sequenece tending to x, the image sequence had *some* convergent subsequence converging ot *something* in the image sequence. It did not show the image sequence converged, or even that the subsequence that converged converged to the right thing.

Others read it as:

"if f is a map of (metric) topological spaces is it true that f(K) compact iff f(K) compact implies f is a homeomorphism"

the answer there is no, since there is nothing ot make f even injective, or surjective, never mind bijective, so we don't even need to consider whether or not f is continuous, or even invertibly continuous.

There are lots of results about in point set topology like this though. For instance every continuous bijection from a compact space to a hausdorff space is a homeomorphism.

Note every metric space is hausdorff, which is one reason why metric spaces are nice topological spaces.