# Non 2nd-countable, Hausdorff, differentiable n-manifold?

1. Nov 3, 2009

### RoNN|3

I am trying to find a Hausdorff topological space that is not second-countable but otherwise a DIFFERENTIABLE n-manifold. I can't figure it out. Does it exist?

I read about the classical example of $$L=\omega_1\times[0,1)$$ with lexicographical order and the order topology. It's Hausdorff, not second-countable and locally homeomorphic to $$\mathbb{R}$$. (found a nice http://www.uoregon.edu/~koch/math431/LongLine.pdf" [Broken]) To make an n-manifold I thought $$L\times[0,1]^{n-1}$$ could work. But is $$L$$ a differentiable manifold? Are the gluing maps $$C^\infty$$? Can the maps be constructed so that they are?

If this doesn't work I have no clue what could it be. Does anyone know an example?

Last edited by a moderator: May 4, 2017
2. Nov 3, 2009

### RoNN|3

Gave it more thought.
To clarify, the definition of an n-differentiable manifold I am using is: 2nd-countable, Hausdorff space $$X$$ s.t. there's an atlas $$\{(U_i,h_i,V_i)\}$$ where $$U_i$$ open in $$X$$, $$\cup{U_i}=X$$, $$V_i$$ open in $$\mathbb{R}^n$$, $$h_i: U_i \rightarrow V_i$$ a homeomorphism and gluing maps are $$C^\infty$$ i.e. $$\forall{i,j}\ { }(h_j\circ{h_i^{-1}}): h_i(U_i\cap{U_j})\rightarrow{h_j(U_i\cap{U_j})}$$ is $$C^\infty$$.

My first idea was to try something with discrete topology but I discarded it when I saw that every book/website sites the Long Line as an example. I thought "It has to be the long line. Why would it exist in the first place if you can do it much easier?!". Anyway, I worked with the first idea a bit and got this:

$$\mathbb{R}^n\times\mathbb{R}$$ with the product topology of standard and discrete. It is Hausdorff and not second-countable. I give the atlas of $$(\mathbb{R}^n\times\{x\}, \pi, \mathbb{R}^n )$$ where $$\pi$$ is projection (it's a homeomorphism). But I don't know if this is valid since there are no gluing maps: all $$U_i$$ and $$U_j$$ are disjoint so this is.. trivial. Does this count??

Now I realize why the long line is always cited: it's path connected. And I don't require any connectedness in my definition. But then this makes one lousy manifold I also remember reading that some definitions exclude spaces with uncountably many connected components. What is the property that guarantees that? If I make my space "countably many copies of $$\mathbb{R}^n$$" I believe it becomes second-countable but I don't want that.

I am confused and frustrated that there are many different definitions for this sort of thing. Wasted a lot of time on this. :grumpy:

3. Nov 4, 2009

### hamster143

One big problem with dropping the 2nd-countable requirement from the definition of a differentiable manifold is that your atlas is no longer required to be countable, that could be inconvenient.

As far as I can tell, it's possible to create an atlas and differentiable gluing maps for the long line in the same manner as the proof of theorem 7 from your link.

4. Nov 5, 2009

### janrozendaal

I think it's a wonderful example of such a manifold. If I guess correctly I know who you are, Ron, since it is a little too convenient that you are trying to find such a space at exactly the same time as I am. So you're probably trying to make the same homework exercises as I am.
I didn't have much time to make them, and was just surfing the net trying to find such a manifold, and the first link I found was your post. It's exactly the answer which I suppose we are to give, it's really nice!
(applause) :)

5. Feb 20, 2011

### Landau

I know this is an old thread, I just happened to come across it.
That's my thought: topological 0-manifolds are the discrete spaces. They are Hausdorff. They are second-countable iff they are countable. So just take an uncountable set with the discrete topology. The atlas is just all singletons, with the unique map into R^0.

Altough you might argue if functions R^0\to R^0 can be called differentiable, i.e. do 0-dimensional smooth manifolds exist?

6. Feb 20, 2011

### Hurkyl

Staff Emeritus
Yes. In fact, off the top of my head I can't think of any construction that doesn't just work as normal in the zero-dimensional case.

As with other finite-dimensional real vector spaces, the tangent space to any (the) point of R0 can be identified with R0 itself.

The tricky question is whether the empty set is a differentiable manifold. (I'm 99.9% sure the answer should be "yes" -- although I would be unsurprised at an author who excludes this degenerate case by fiat)

7. Feb 20, 2011

### Landau

Sure. My problem is earlier: a function M->R is called smooth if for every chart (U,\phi) the composition $f\circ\phi^{-1}:\phi(U)\to\mathbb{R}$ is smooth. But in the case of a 0-manifold,
the domain $\phi(U)=\{p\}=\mathbb{R}^0$ is a singleton. As I said, I am wondering if such a map may be called differentiable. Usually in real analysis one only defines the derivative for functions $\mathbb{R}^{n>0}\to\mathbb{R^m}$.

8. Feb 20, 2011

### Hurkyl

Staff Emeritus
Yes, every function from R0 to itself is differentiable.

There are several ways to define differentiable functions on Euclidean space and the category of differentiable manifolds that I actually forgot about the limit of (norms of) difference quotients.

9. Feb 20, 2011

### Landau

I am more interested in an explanation. Of course we could just require this by definition, but I am wondering whether there is a natural reason for this.

R0 is a 0-dimensional real vector space. The usual definition reads: f:R0->Rn is differentiable at p if there exists a linear map L:R0->Rn such that

$$\lim_{x->p}\frac{\|f(x)-f(p)-L(x-p)\|}{\|x-p\|}=0.$$

Now p is the unique point in R0, so L must be the zero linear map and f(x)-f(p)=0=L(x-p), and x=p. We are forced to define the norm on R0 to be zero. So this whole definition does not make any sense.

10. Feb 20, 2011

### mathwonk

since you did not say it had to be connected the answer is trivial.

11. Feb 21, 2011

### hamster143

By this definition, the limit does not exist, because p is not a limit point of this vector space. R0 has no limit points at all. Therefore, functions from R0 to R0 can't be differentiable. (Unless you explicitly declare them differentiable as a special case, of course.)

On the other hand, any mapping from R0 to R0 is an identity mapping, and identity mappings are usually differentiable everywhere.

Last edited: Feb 21, 2011
12. Feb 21, 2011

### Hurkyl

Staff Emeritus
Oh! Now that I pay attention, I realize that this definition really does make sense.

If you pay careful attention to the definition of the limit, you find that
$$\lim_{x \to p} f(x) = a$$​
is true for every value of a and for every function f, if the domain is the one-point space. This is a fact that somehow escaped my attention until now. Note that this isn't nonsense -- instead it means that the definition limit only defines a relation instead of a partial function in f and p as it usually does.

If you pay careful attention to the definition of linear map, you'll find that the zero map is the only linear map whose domain is the zero-dimensional vector space.

The norm on R0 is zero, by the way.

The "natural reason" is that everything you want to do with derivatives works exactly as they should. The Jacobian formula for the total derivative makes sense and gives the right (and only) result. You have the Taylor series
$$f(x + \epsilon) = f(x) + f'(x) \epsilon + f''(x) \epsilon^2 / 2 + \cdots$$​

Geometrically, if two one-dimensional differentiable manifolds don't intersect in a terrible fashion, then their intersection is a collection of zero-dimensional differentiable manifolds (or the empty differentiable manifold)

The tangent space to a point is just a point.

And so on, and so on. Try thinking of anything you think ought to be true about derivatives -- you'll see that it is still true when the domain is the one-point space.

Last edited: Feb 21, 2011
13. Feb 21, 2011

### Landau

True, IF you require p to be a limit point in your definition of limit. I know this is usually done, but not always. (e.g. in my introductory analysis class we did not look at punctured neighborhoods of p, but just neighborhoods of p).

Again, this is rather dependent on your precise definition of 'limit'. Different people may have subtle differences in their definitions, and in subtle cases like these it is advisable to be explicit about them.

First, p is not a limit point, as hamster noted. So if this is a requirement of 'our' definition of limit, then your claim is invalid. Second, in my previous post the limit is taken for a certain quotient: for the 1-point space this quotient is undefined, since the denominator is zero. So we are not in the case of

$$\lim_{x \to p} f(x) = a$$​

since our f, the quotient of norms

$$x\mapsto \frac{\|f(x)-f(p)-L(x-p)\|}{\|x-p\|}=\frac{0}{0}$$

is not defined at all. So even if p does not need to be a limit point, I don't see how your claim is relevant.
That's what I already said (and it is trivial: zero must go to zero, and there is no other vector).
That's what I already said (and it is trivial: zero must have zero norm, and there is no other vector).
I am still not convinced about the definition of the total derivative from my previous post.

Last edited: Feb 21, 2011
14. Feb 21, 2011

### Hurkyl

Staff Emeritus
That's fair -- I don't think I'd seen "p is a limit point of some subset of the domain" before.

This one is a problem even in positive dimension, and is dealt with in a variety of standard ways, such as plugging in an arbitrary value to fill the hole, or only requiring the argument of the limit to be defined on a punctured neighborhood.

And your difference quotient is defined on a punctured neighborhood of p -- your variable x ranges over the empty set, and so the image of ||x - p|| is a subset of the nonzero reals.

The way you said it, you made it sound like it was the final step in deriving an absurdity.

15. Feb 21, 2011

### Landau

I suspect you have: that way we can sensibly talk about the limit by considering the function only on a punctured neighborhood (an approach which you just mentioned). If p is not a limit point, some of those punctured neighborhoods is empty, hence the image of such a punctured neighborhood is empty. See below.
Yes, but those 'standard ways' are assuming that p is a limit point! As for the punctured neighborhood solution if p is not a limit point, see below. Plugging in an arbitrary value to fill the hole seems pretty ugly, since in this case there is nothing but a hole! Doing this is exactly the same as just postulating that the limit exists.
Yes, that image is empty. And then? Are you suggesting the following definition+reasoning?

Definition: Suppose $f:\mathbb{R}^{n\geq 0}\to\mathbb{R}^{m\geq 0}$. Let $p\in\mathbb{R}^{n}$ be arbitrary (so not necessarily a limit point!). We say $\lim_{x\to p}f(x)=b\in\mathbb{R}^m$ when for every e>0 there exists d>0 such that for all $x\in\mathbb{R}^n$:

$$0<\|x-p\|<d\Rightarrow \|f(x)-b\|<e.$$

(0<|x-p|, so consider only punctured neighborhoods!)

Reasoning: In the case $n=0, \mathbb{R}^0=\{p\}, f:\{p\}\to R^m$ any function, the implication holds vacuously. Hence f converges to any b.

That's not very nice: any f converges to any b. So we don't even have unique limits.

Of course, the same is true whenever p is not a limit point. In fact, in my introductory analysis class we used this definition, but the concept of 'limit point' was motivated precisely by the fact that limits are not unique (at all) in this case. It was remarked that for most practical purposes this behaviour was not interesting. But since we are discussing the one-point space in this thread, we need to be careful about this.

Yes, I should have phrased it differently.

Last edited: Feb 21, 2011
16. Feb 21, 2011

### hamster143

The limit point requirement is there for a good reason.

Consider the function:

f(x) = sqrt(x) for x>=0
f(x) = 0 for x=-1

If we drop the limit point requirement, then f is continuous and differentiable at x=-1 (using the topological definition of the limit, and the topology induced on its domain by the topology of the real line).

Last edited: Feb 21, 2011
17. Feb 21, 2011

### Landau

I think my illustration of what happens if we drop the limit point requirement is better: your f converges to any point if x tends to -1. I don't really mind your f being continuous at -1.

18. Feb 23, 2011

### jasomill

The appendix to volume 1 of Spivak's Comprehensive Introduction to Differential Geometry has several nice examples of non-metrizable manifolds.

As for calculus in $$\mathbb{R}^0$$, the (trivial) problems vanish if we define things as follows: say a map $$f: X \to Y$$, where X and Y are (possibly trivial) Banach spaces, is differentiable at $$x \in X$$ iff there exists a (continuous*) linear map $$L: X \to Y$$ such that

$$f(x + h) = f(x) + Lh + \left\Vert{h}\right\Vert\varphi(h)$$

for some map $$\varphi$$ such that $$\varphi(h) \to 0$$ as $$\left\Vert{h}\right\Vert \to 0$$.

If X or Y is trivial, f is constant, thus differentiable — L must be zero in any case, and $$\varphi = 0$$ works.

Definition of "limit point" notwithstanding, how could any sequence in a topological space with fewer than two points fail to converge?

Cheers,
Jason

* Not really relevant here, since constant maps are continuous and trivial vector spaces are finite-dimensional; just covering the bases!