To infinity and beyond

[Ben]

Anyway kids, now we are ready to tackle parallelism head on.

 

Suppose I take a segment [math]A[/math] of the real line and join the two ends together. This we recognize as a circle; grand folk call it the 1-sphere [math]S^1[/math]. Now since this is essentially the "same" as our line segment [math]A[/math], we might expect to be able to find a one-to-one correspondence between them. This correspondence - an isomorphism on sets - will be given by an invertible set function from one to the other (this is called a bijection, btw).

 

Of course no such function exists; the map [math]A \to S^1[/math] sends the two ends of [math]A[/math] to same point of [math]S^1[/math] - our "join". And since well-behaved functions are not allowed to send a single point in the domain to distinct points in the co-domain, there can be no inverse [math]S^1 \to A[/math].

 

Now there is a theorem (due I think to Dedekind and Pierce) that states that an infinite set is always isomorphic to a proper subset of itself. Since [math]A \subsetneq R^1[/math] and since isomorphism is an equivalence relation, by transitivity we might expect to be able to do something like the following.

 

Place our circle anywhere on the real line with our join at the North pole, and project each point of [math]S^1[/math] "down" to a point in [math]R^1[/math] in such a way that none of these projections intersect each other or [math]S^1[/math].

 

All goes swimmingly until we get to the North pole, where now our projections run parallel to [math]R^1[/math] and seem never to meet it.

 

But we have the perfect construction specifically made for just this purpose, called the "projective real line". This is formed by adjoining what's called the "point at infinity" to [math]R^1[/math] to form [math]R^1 \cup \{\infty\}[/math].

 

So now our point at the North pole projects to this point at infinity, and an isomorphism [math]S^1 \simeq R^1 \cup \{\infty\}[/math] can be recovered, but at some cost; parallel lines meet!

 

Worse still, since the two projections - "left" and "right" - from the North pole map to the same point in our projective line, this implies that, in [math]R^1 \cup \{\infty\}[/math], the "end-points" of the real line [math]R^1[/math] may be considered as the same point.

 

PS. Don't blame me if you don't like it - it is mainstream mathematics

[Qfwfq]

Don't blame me if you don't like it - it is mainstream mathematics

I like it. ;)

 

How about some fun with bijectively (perhaps even homeomorphically) mapping [imath]\mathbb{R}^2[/imath] and [imath]\mathbb{R}[/imath]? :)

[freeztar]

I think I understand most of your last post, Ben, but I'm stuck on this part:

 

So now our point at the North pole projects to this point at infinity, and an isomorphism [math]S^1 \simeq R^1 \cup \{\infty\}[/math] can be recovered, but at some cost; parallel lines meet!

 

Can you elaborate on this part please? Keep in mind that my math skill level is relatively low. So, no need to feel that you might be insulting me by "dumbing it down" too much. ;)

[Ben]

OK, it's largely a matter of notation and definition, like......

 

If I write [math] (0,1)[/math] as a subset of the real line [math]R^1[/math], I simply mean it is the "interval" that contains all the real numbers between 0 and 1, but not zero or 1 themselves.

 

It is customary, since we have no way of knowing "how far" the real line extends, to write it long-hand as [math](-\infty, \infty)[/math] - no infinities here, then.

 

The notation [math]\{x\}[/math] denotes what is called a "singleton set"; this is a set with a single member. The slightly annoying thing is that this is not at all the same thing as [math]x[/math] itself (there is a good reason for this, but requires a bit of set theory).

 

The "union" of the sets [math]A,\,\,B[/math] is the set that contains all elements of [math]A[/math] and all those of [math]B[/math], where by definition if these 2 sets contain an element in common, they must not be counted twice.

 

The projective line [math]R^1 \cup \{\infty\}[/math], a different beast entirely, is thus the union of a set that doesn't contain an element called "infinity" with a singleton set that is called the point at infinity.

 

The consequences are as I tried to describe

[Ben]

How about some fun with bijectively (perhaps even homeomorphically) mapping [imath]\mathbb{R}[/imath] and [imath]\mathbb{R}[/imath]?

I agree this would be fun, but I'm not sure it would be relevant to this thread. I could ramble on at length on this subject - it is one my faves, but I should be very embarrassed to open yet another thread on this forum, so how about you give it a shot?

[Qfwfq]

I'm not sure it would be relevant to this thread.

The existence of bijective maps shows they have the same cardinality so I suggested it because it seemed relevant.

 

I notice now that I had left out the square! :eek: I meant the real line and the plane, of course. I'm no expert on such mappings, I was hoping you'd show us some tricks. I know bijective ones exist, I doubt they can also be homeomorphic (for their usual topologies) but I'm not 100% sure if it's formally impossible.

[freeztar]

OK, it's largely a matter of notation and definition, like......

 

If I write [math] (0,1)[/math] as a subset of the real line [math]R^1[/math], I simply mean it is the "interval" that contains all the real numbers between 0 and 1, but not zero or 1 themselves.

 

It is customary, since we have no way of knowing "how far" the real line extends, to write it long-hand as [math](-\infty, \infty)[/math] - no infinities here, then.

 

The notation [math]\{x\}[/math] denotes what is called a "singleton set"; this is a set with a single member. The slightly annoying thing is that this is not at all the same thing as [math]x[/math] itself (there is a good reason for this, but requires a bit of set theory).

 

The "union" of the sets [math]A,\,\,B[/math] is the set that contains all elements of [math]A[/math] and all those of [math]B[/math], where by definition if these 2 sets contain an element in common, they must not be counted twice.

 

The projective line [math]R^1 \cup \{\infty\}[/math], a different beast entirely, is thus the union of a set that doesn't contain an element called "infinity" with a singleton set that is called the point at infinity.

 

The consequences are as I tried to describe

 

Thanks for that.

 

I'm still confused by this statement though:

 

an isomorphism S^1 \simeq R^1 \cup \{\infty\} can be recovered, but at some cost; parallel lines meet!

[Ben]

The existence of bijective maps shows they have the same cardinality

Umm, not sure what you mean here. I think perhaps it's the other way round... only sets of the same cardinality may have bijections between them

 

Like... suppose that [math]S,\,\,T[/math] are sets of the same cardinality. Let this cardinality be countable, possibly infinitely so.

 

Then it is easy-peasy to show they are are isomorphic: we have that [math]S \simeq \mathbb{N}[/math] and [math]T \simeq \mathbb{N}[/math]. And since isomorphism is an equivalence relation, we may assume symmetry and transitivity, so [math] S \simeq T[/math]. Which of course implies the existence of a bijection.

 

However, where our sets are uncountable, we have a little more work to do, namely define invertible functions first. If we are to carry even one more forum member along with us, it will require some explanation (for which I have no time right now - leave it with me a day).

 

the real line and the plane I doubt they can also be homeomorphic (for their usual topologies)

This will require even more explanation, but I think your "doubts" are unfounded. Taster: the product of homeomorphisms is a homeomorphism on products.

 

Exercise for the reader (teasing)

 

I am decorating my "study" here, so my texts are boxed away, so lemme think of a way to make it clear.

[Qfwfq]

Gosh it seems we're into a bit of misunderstanding here. Maybe you read my post in the midst of your redecoration work! When you've fully recovered from it, show us some examples of mappings:

 

[math]\varphi:\,\mathbb{R}^2\rightarrow\mathbb{R}[/math]

 

that are bijective and then discuss whether there can be homeomorphic ones.

[Ben]

OK, so the issue before us is this: does there exist a homeomorphism between [math]\mathbb{R}[/math] and [math]\mathbb{R}^2 \equiv \mathbb{R} \times \mathbb{R}[/math]?

 

So the first thing I want to say is that the real numbers are exceptionally irritating to deal with. Why? Because we can treat them as a set, as a vector space, as a topological space, as a manifold...... This would be OK, if we ALL were not, on occasion, tempted to forget how we were thinking of them.

 

In the present case, we will take the real numbers first to be a set, and then as a topological space. I will try to emphasize the distinction as we go.

 

Second is that, as has been pointed out to me, occasionally somewhat tartly, that "infinity" is not a number, or a place or anything like that.

 

Third, I trust we all remember the difference between an ordinal number and a cardinal number - the first in some loose sense describes the "position" of an object in a set, the second describes the "size" of that set.

 

So. Recall in my OP that I claimed, essentially, that the (infinite) cardinality of the natural numbers [math]\mathbb{N}[/math] is strictly less than that of its powerset [math]\mathcal{P}\mathbb{N}[/math] and posed the question: what does it mean to say that [math]\infty < 2^{\infty}[/math]?

 

Let me recant. We will call the cardinal number associated to the set of natural numbers as [math]\aleph_0[/math]. This is its cardinality. Then, to the powerset [math]\mathcal{P}\mathbb{N}[/math] I can associate a cardinal number defined as [math]2^{\aleph_0} = \mathfrak{c}[/math].

 

Now the same reasoning as in the OP can be used to show that the cardinality of the SET of real numbers is also [math]\mathfrak{c}[/math] - it's called the "cardinality of the continuum" (don't ask me why).

 

So let's quickly do this: if the cardinality of [math]\mathbb{R} = \mathfrak{c}=2^{\aleph_0}[/math], then the cardinality of [math]\mathbb{R} \times \mathbb{R} = 2^{\aleph_0} \times 2^{\aleph_0} = 2^{\aleph_0 + \aleph_0} = 2^{\aleph_0} = \mathfrak{c}[/math].

 

So that [math]\mathbb{R}[/math] and [math]\mathbb{R} \times \mathbb{R}[/math], have the same cardinality namely [math]\mathfrak{c}[/math], and are therefore isomorphic, essenially by definition; the bijection on sets [math] f:\mathbb{R} \to \mathbb{R} \times \mathbb{R}[/math], with inverse, can be assumed to exist.

 

Question now is: when we consider the real numbers and their Cartesian product as topological spaces, does this bijection a.k.a. homeomorphism, exist? Of course, for this we need to establish continuity in "both directions".

 

Later

 

P.S Posts crossed!

[Qfwfq]

Now the same reasoning as in the OP can be used to show that the cardinality of the SET of real numbers is also [math]\mathfrak{c}[/math] - it's called the "cardinality of the continuum" (don't ask me why).

I think it's cuz the reals are considered to constitute a continuum, unlike the rationals (which seems counterintuitive). The rationals are dense in the reals, between any two of them there are irrationals; the reals are complete, there can't be "irreals in between them" so they're said to be a continuum.

 

So that [math]\mathbb{R}[/math] and [math]\mathbb{R} \times \mathbb{R}[/math], have the same cardinality namely [math]\mathfrak{c}[/math], and are therefore isomorphic, essenially by definition; the bijection on sets [math] f:\mathbb{R} \to \mathbb{R} \times \mathbb{R}[/math], with inverse, can be assumed to exist.

I'm with you up to here and I was hoping you might be familiar with a few cool examples of bijections. I know many exotic generalized curves have been defined, each of which "fills" [imath]\mathbb{R}^2[/imath] without crossing itself, but I don't know the details much.

 

Question now is: when we consider the real numbers and their Cartesian product as topological spaces, does this bijection a.k.a. homeomorphism, exist? Of course, for this we need to establish continuity in "both directions".

Whether a bijection is also a homeomorphism depends on which topology is given to each set. It can always be made so by inducing one of the two topologies from the other, via the bijection, but this wouldn't guarantee both of them being the ordinary ones of the line and the plane (induced by the metric). Much more likely they wouldn't both be.

 

If it were somehow possible, it would clash with the necessity of using more than one coordinate to describe the geometry of the plane; n-dimensional Euclidean spaces would come across as a hoax, but I was curious to see how the impossibility could be directly proven.

 

Edit: I mean proven formally, it seems obvious enough intuitively.

[Ben]

I know many exotic generalized curves have been defined, each of which "fills" [imath]\mathbb{R}^2[/imath] without crossing itself, but I don't know the details much.

Hmm, the only such that am familiar with are the integral curves. These, of course, require us to think of [math]\mathbb{R}^n[/math] as a manifold; roughly speaking, these are the curves that "follow" one vector in a tangent space to another vector in another such space. Is that what you were referring to? probably not.

 

If it were somehow possible, it would clash with the necessity of using more than one coordinate to describe the geometry of the plane; n-dimensional Euclidean spaces would come across as a hoax, but I was curious to see how the impossibility could be directly proven.

Actually I now see you were right, and I was wrong; there can be NO homeomorphism [math]\mathbb{R} \simeq \mathbb{R}^2[/math]

 

But lookee here. I am busy packing up to go on 2 week's vacation, and don't have a lot of time to write out a formal proof, so I will give a quick-and-dirty.

 

Suppose the homeomorphism [math]\mathbb{R} \simeq \mathbb{R}^2[/math].

 

This means they are "topologically equivalent". Then, since [math]\mathbb{R}^2 \equiv \mathbb{R} \times \mathbb{R}[/math], then what ever I do to [math]\mathbb{R}[/math] must carry over into [math]\mathbb{R}^2[/math], while maintaining this equivalence.

 

So let's remove the origin from [math]\mathbb{R} \mapsto \mathbb{R}\setminus \{0\}[/math], and then form [math]\mathbb{R}\setminus\{0\} \times \mathbb{R} \setminus \{0\} \equiv \mathbb{R}^2\setminus \{0,0\}[/math].

 

Now connectedness is a topological property that must be preserved under homeomorphism. But, [math]\mathbb{R}^2\setminus \{0,0\}[/math] is connected, whereas [math]\mathbb{R}\setminus \{0\}[/math] is not.

 

Thus, as topological spaces, [math]\mathbb{R} \not\simeq \mathbb{R}^2[/math].

 

Or something like that.

 

So, guys, play nice while I'm away, OK, or Daddy will be cross

[Qfwfq]

...don't have a lot of time to write out a formal proof, so I will give a quick-and-dirty.

Actually that's quite a good argument, quick but not dirty. I was trying to think of the topological properties of neighborhoods of an arbitrary point being conserved but I didn't think of removing the point. ;)

 

So, guys, play nice while I'm away, OK, or Daddy will be cross

The mice will be dancing! ;)