To infinity and beyond

Ben · 2 Weeks Ago

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

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 $\mathbb{N}$ is strictly less than that of its powerset $\mathcal{P}\mathbb{N}$ and posed the question: what does it mean to say that $\infty < 2^{\infty}$ ?

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

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

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

So that $\mathbb{R}$ and $\mathbb{R} \times \mathbb{R}$ , have the same cardinality namely $\mathfrak{c}$ , and are therefore isomorphic, essenially by definition; the bijection on sets $f:\mathbb{R} \to \mathbb{R} \times \mathbb{R}$ , 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 · 2 Weeks Ago

Quote:

Originally Posted by Ben

Now the same reasoning as in the OP can be used to show that the cardinality of the SET of real numbers is also $\mathfrak{c}$ - 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.

Quote:

Originally Posted by Ben

So that $\mathbb{R}$ and $\mathbb{R} \times \mathbb{R}$ , have the same cardinality namely $\mathfrak{c}$ , and are therefore isomorphic, essenially by definition; the bijection on sets $f:\mathbb{R} \to \mathbb{R} \times \mathbb{R}$ , 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" $\mathbb{R}^2$ without crossing itself, but I don't know the details much.

Quote:

Originally Posted by Ben

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 · 2 Weeks Ago

Quote:

Originally Posted by Qfwfq

I know many exotic generalized curves have been defined, each of which "fills" $\mathbb{R}^2$ 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 $\mathbb{R}^n$ 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.

Quote:

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 $\mathbb{R} \simeq \mathbb{R}^2$

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 $\mathbb{R} \simeq \mathbb{R}^2$ .

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

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

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

Thus, as topological spaces, $\mathbb{R} \not\simeq \mathbb{R}^2$ .

Or something like that.

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

Qfwfq · 2 Weeks Ago

Quote:

Originally Posted by Ben

...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.

Quote:

Originally Posted by Ben

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

The mice will be dancing!

To infinity and beyond

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