Connected metric space problem

[radou]

Homework Statement

One needs to show that a connected metric space having more than one point is uncountable.

The Attempt at a Solution

First of all, if (X, d) is a connected metric space, it can't be finite, so assume it's countably infinite. Let x be a fixed point in X. For any x1 in X different from x, there exist disjoint open neighborhoods V of x1 and U1 of x. Since they are disjoint, Cl(U1) does not contain x1. The intersection of Cl(Ui) is closed and contains the element x. Now, if we define a function f : N --> X with f(n) = xn, then f must be surjective, but x does not equal f(n) for any n, since for any n xn is not in Cl(Un). Hence, X is uncountable.

I'm not really sure about this, so any help is appreciated.

 

[micromass]

Hmmm, firstly you say that the map f is surely surjective, why is that?
Second, you never used the assumption that X is connected, I think.

As a hint. Can you find a function from X to [0,1] which is non-trivial? Think Urysohns lemma. Have you seen this?

 

[Dick]

Yes, the 'proof' is bogus. Take X to be the rationals Q. Is then anything in your 'proof' that stops you from PROVING the rationals are uncountable? If not, then it's WRONG. Now try and figure out why. To give you a hint in the positive direction (and I don't think you need Urysohn's lemma), suppose x and y are two points in your metric space and d(x,y)=c. Define S(x,r)={z:d(x,z)=r} and suppose there is a number 0<e<c such that S(x,e) is empty. Think about it.

 

[micromass]

Or you could just take two point x and y and define the function

$$f:X\rightarrow [0,1]:z\rightarrow \frac{d(z,x)}{d(z,x)+d(z,y)}$$

And then use that the image of a connected set is connected...

 

[Dick]

Or you could just take two point x and y and define the function

$$f:X\rightarrow [0,1]:z\rightarrow \frac{d(z,x)}{d(z,x)+d(z,y)}$$

And then use that the image of a connected set is connected...

For one thing why do you think your f takes X->[0,1]? It doesn't. And for another, I AM suggesting using that the image of a connected set is connected. I'm just suggesting radou prove it in this specific context.

 

[radou]

The Attempt at a Solution

First of all, if (X, d) is a connected metric space, it can't be finite, so assume it's countably infinite. Let x be a fixed point in X. For any x1 in X different from x, there exist disjoint open neighborhoods V of x1 and U1 of x. Since they are disjoint, Cl(U1) does not contain x1. The intersection of Cl(Ui) is closed and contains the element x. Now, if we define a function f : N --> X with f(n) = xn, then f must be surjective, but x does not equal f(n) for any n, since for any n xn is not in Cl(Un). Hence, X is uncountable.

I'm not really sure about this, so any help is appreciated.

OK, the mistake in this proof is that f needn't be surjective. If a set X is countably infinite, that only tells us that there exists a surjection f : N --> X, nothing else.

I'll think about what you wrote down.

 

[Dick]

OK, the mistake in this proof is that f needn't be surjective. If a set X is countably infinite, that only tells us that there exists a surjection f : N --> X, nothing else.

I'll think about what you wrote down.

Right. It just tells you there exists a surjection. It DOESN'T tell you that if a specific map isn't a surjection that the set is uncountable. If X is countable there are still LOTS of maps from N -> X that aren't surjections. As you just showed.

 

[radou]

Right. It just tells you there exists a surjection. It DOESN'T tell you that if a specific map isn't a surjection that the set is uncountable. If X is countable there are still LOTS of maps from N -> X that aren't surjections. As you just showed.

OK, thanks. One of my reckless mistakes again, as usual.

OK, if I got it right, the strategy of your hint for the proof is to assume such an e exist, and if we arrive at a contradiction, the negation of the statement must hold.

So, assume there is some 0 < e < c such that S(x, e) is empty. Then X can be represented as the union of disjoint, non-empty and open sets B(x, e) = {z in X : d(x, z) < e} and B' = {z in X : d(x, z) > e}, contradicting the fact that X is connected.

Hence, for any e > 0, the set S(x, e) is non-empty, and for any e1 and e2 > 0, these sets are disjoint. Now, since <0, +∞> is uncountable, X must be uncountable.

Does this work?

 

[Dick]

OK, thanks. One of my reckless mistakes again, as usual.

OK, if I got it right, the strategy of your hint for the proof is to assume such an e exist, and if we arrive at a contradiction, the negation of the statement must hold.

So, assume there is some 0 < e < c such that S(x, e) is empty. Then X can be represented as the union of disjoint, non-empty and open sets B(x, e) = {z in X : d(x, z) < e} and B' = {z in X : d(x, z) > e}, contradicting the fact that X is connected.

Hence, for any e > 0, the set S(x, e) is non-empty, and for any e1 and e2 > 0, these sets are disjoint. Now, since <0, +∞> is uncountable, X must be uncountable.

Does this work?

It's close. The set [0,c] is uncountable. And I would say for any e in [0,c] (not any e>0), S(x,e) is nonempty. Can you use that to define a surjection of X onto [0,c]?

 

[radou]

Hence, for any e > 0, the set S(x, e) is non-empty, and for any e1 and e2 > 0, these sets are disjoint. Now, since <0, +∞> is uncountable, X must be uncountable.

Does this work?

It's close. The set [0,c] is uncountable. And I would say for any e in [0,c] (not any e>0), S(x,e) is nonempty. Can you use that to define a surjection of X onto [0,c]?

OK, I'll first correct my argument above.

For any e in [0, c] (I overlooked the assumption that e < c in the beginning and that for e = 0 S(x, e) = {x}, for some weird reason at a point, I mistook the S's for open balls), the set S(x, e) is non-empty.

Now, for any e in [0, e] let z be the element of S(x, e) such that d(z, y) = inf{d(x, y)} : x is in S(x, e)}. This seems well defined, since S(x, e) is closed (its complement is open), so the distance from y to S(x, e) equals d(z, y), for some z in S(x, e). Now we have a bijective correspondence from [0, c] onto a subset of X (if we set the codomain equal the image), hence this subset is uncountable. It seems correct to conclude that X is uncountable too, since it contains an uncountable subset, but I'm not really sure about this.

 

[micromass]

Yes, every set which contains an uncountable set is uncountable to. Methinks you've got it...

For your convenience, I'll also present the argument which I had in mind.

Take two different point x and y, then define the function

$$f:X\rightarrow \mathbb{R}:z\rightarrow \frac{d(z,x)}{d(z,x)+d(z,y)}$$

This function has f(x)=0 and f(y)=1. Since X is connected, we have that f(X) is connected. Since 0 and 1 are in f(X). We must have that f(X) contains [0,1]. Thus f is a surjection onto [0,1]. This means that X has to be uncountable.

But I like the other proof better

 

[radou]

You mean f is a surjection onto a set which contains [0, 1]? Or am I missing something.

 

[micromass]

Well, therange of f is [0,1], so it is a surjection onto [0,1].

But knowing that the range contains an uncountable set seems to be enough to...

 

[radou]

Hm, it confuses me that Dick claims the opposite in post #5 about the range of f.

 

[micromass]

Yeah, that surprises me to.

f is clearly positive, since all metrics are possitive.

We also have that

$$0\leq d(z,y)$$

hence

$$d(z,x)\leq d(z,x)+d(z,y)$$

thus

$$\frac{d(z,x)}{d(z,x)+d(z,y)}\leq 1$$

So $$f(z)\in [0,1]$$. I don't really see Dicks problem...

 

[radou]

We could define this function as $$z\rightarrow \frac{d(z,x)}{d(y,x)+d(z,y)}$$ too, right? The triangle inequality assures that d(z, x) <= d(x, y) + d(y, z), and f(x) = 0, f(y) = 1 holds, too.

 

[micromass]

Yes, that's a good function to.

This kind of function is very important in topology, it can be used to show many great things about metric spaces. It's called a Urysohn function.

 

[Dick]

Hm, it confuses me that Dick claims the opposite in post #5 about the range of f.

Don't worry about it. I misread the function.

 

[radou]

OK, thanks to all for the help!