
#1
Feb613, 03:50 PM

P: 23

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 



#2
Feb613, 03:57 PM

Sci Advisor
HW Helper
PF Gold
P: 2,897

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})##.




#3
Feb713, 02:54 AM

P: 23

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? 



#4
Feb713, 02:57 AM

Mentor
P: 16,542

What is {ℝ}???
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. 



#5
Feb713, 05:32 AM

Sci Advisor
P: 773

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



#6
Feb713, 09:26 AM

P: 23

@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. 



#7
Feb813, 07:00 AM

Sci Advisor
P: 773

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. 



#8
Feb913, 10:30 AM

P: 23

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



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