What is {ℝ}???

kokolovehuh

Hi,
Someone I know tried to convey me the meaning of {[itex]ℝ[/itex]}, stating it represents a set of real numbers. But using notation, {[itex]ℝ[/itex]}, is implying that the real space is (improperly) contained in a set, and I don't think this makes any logical sense.
On the other hand, we can say {[itex]x \in S | \forall S \in ℝ[/itex]}, etc....or simply [itex]x \in ℝ[/itex].
Another way of thinking about this is instead of putting your foot in a sock, you are putting the sock into your foot and it's disturbing.

Am I right or wrong?

Thanks

jbunniii

It is perfectly valid to put sets inside of other sets. For example, we may consider the set of all subsets of the complex numbers ##\mathbb{C}##. This is called the power set of ##\mathbb{C}## and is sometimes given the notation ##\mathcal{P}(\mathbb{C})## or ##2^\mathbb{C}##. Any subset of ##\mathbb{C}## is an element of ##\mathcal{P}(\mathbb{C})##. For example, we have ##\mathbb{R} \subset \mathbb{C}## and ##\mathbb{R} \in \mathcal{P}(\mathbb{C})##. We can form subsets of ##\mathcal{P}(\mathbb{C})## in the usual way, by putting elements of ##\mathcal{P}(\mathbb{C})## into a set. Thus ##\{\mathbb{Z}, \mathbb{Q}, \mathbb{R}\} \subset \mathcal{P}(\mathbb{C})##, and a special case is a subset containing only one set, such as your example: ##\{\mathbb{R}\} \subset \mathcal{P}(\mathbb{C})##.

kokolovehuh

Thanks bjunniii, you have a good point. However, let me rephrase my doubt.

We want to use a notation to represent a set of all real number, say [itex]X[/itex]. It is immediately apparent that [itex]x \in X \subseteq ℝ[/itex] for some real number [itex]x[/itex]. In this case, we are not not considering any stronger set, for instance, [itex]P(ℂ)[/itex] as you mentioned.
Now having limited ourselves to real space, it is rather redundant to say the set is represented as {[itex]ℝ[/itex]} because since [itex]ℝ[/itex] is not a proper subspace in this case. This is the reason why I said "this does not make any logical sense"; I am ridiculed by those curly brackets!

Instead, we could simply write [itex]X \in ℝ[/itex] that give a much more direct and sensible idea of what space we are talking about.

Do you agree?

micromass

The set [itex]\{\mathbb{R}\}[/itex] is a set which contains only one element. Its element is [itex]\mathbb{R}[/itex]. There is no reason why such a construction would not be allowed.
It is true, however, that sets like [itex]\{\mathbb{R}\}[/itex] don't play a big role in mathematics.

pwsnafu

Unless you state otherwise, all set theory is done in ZFC, and one of its axioms is the axiom of pairing. It asserts: given any two sets A and B, there exists set C with exactly those two elements, i.e. C = {A, B}.

So it does logically exist (in this case A = B = ℝ).

kokolovehuh

@micromass, @pwsnafu, I see what you are saying. Overall, you have convinced me {[itex]ℝ[/itex]} is possible.

But, The notation with curly bracket is directly defining the single element in this set as the real space which is essentially another set. I was simply saying there is no necessity to put real space as a subset of a set in the first place; there are no other disjoint elements.

pwsnafu

I was about to write "what's your point?" until I re-read the original post:

Your friend is wrong. The set of real numbers is ℝ. (I bold "the" because there is only one.)

But ##\{\mathbb{R}\} \neq \mathbb{R}##. The left hand side is "a set with one element, and that element is ℝ". They are not the same thing.

kokolovehuh

Aha, there we go. Thanks alot pwsnafu for the verification! as well as others for your time generous reply.