Cardinality of infinite sequences of real numbers

Click For Summary
SUMMARY

The cardinality of the set of infinite sequences of real numbers, denoted as |ℝ^ℕ|, is equal to the cardinality of the set of real numbers, |ℝ|. The proof involves defining two injective maps: one from |ℝ^ℕ| to |ℝ| and another from |ℝ| to |ℝ^ℕ|. A key aspect of the proof is recognizing that elements of the Cartesian product |[0,1)^2| can be represented as pairs of real numbers with infinite decimal expansions, which can then be combined into a unique real number. Care must be taken to ensure that each real number is represented by a non-terminating decimal expansion to avoid ambiguity.

PREREQUISITES
  • Understanding of cardinality and its implications in set theory
  • Familiarity with real numbers and their properties, particularly decimal expansions
  • Knowledge of injective functions and bijections in mathematics
  • Basic concepts of sequences and functions, especially in the context of countable sets
NEXT STEPS
  • Study the properties of infinite sequences and their cardinalities in set theory
  • Learn about the Cantor-Bernstein-Schröder theorem and its applications
  • Explore the concept of transfinite sequences and their implications in advanced mathematics
  • Investigate the theory of nets and their role in topology and analysis
USEFUL FOR

Mathematicians, students of advanced mathematics, and anyone interested in set theory, particularly those exploring the properties of infinite sets and cardinality.

Berrius
Messages
19
Reaction score
0
I have to prove that the cardinality of the set of infinite sequences of real numbers is equal to the cardinality of the set of real numbers. So:
A := |\mathbb{R}^\mathbb{N}|=|\mathbb{R}| =: B

My plan was to define 2 injective maps, 1 from A to B, and 1 from B to A.
B <= A is trivial, just map a real number x on the sequence (xxxxxxxxx...). But I can't find a injective map from A to B. Can someone help?
 
Physics news on Phys.org
You could reduce it to an easier problem first, for example. Something like |[0,1]^\mathbb{N}|=|[0,1]|
 
I would start by thinking about why the cardinality of ##[0,1)^2## is equal to the cardinality ##[0,1)##. To do this you think realize that any element of ##[0,1)^2## can be written as ##(x,y)## where ##x## and ##y## have infinite decimal expansions ##x = a_1 a_2 a_3 ...## and ##y = b_1 b_2 b_3 ... ##, then you can combine these into a unique real number ##z = a_1 b_1 a_2 b_2 a_3 b_3 ...## .

From here, you can generalize this proof to show that ##|[0,1)^\mathbb{N}| = |[0,1)|## by recalling the proof that the rational and natural numbers have the same cardinality. At this point you should be almost home.

Good Luck!
 
Last edited:
hapefish said:
I would start by thinking about | \mathbb(R) ^{2} | = | \mathbb{R} |. it is much easier to think about |[0,1)^2| = |[0,1)|. To do this you think realize that any element of [0,1)^2 can be written as (x,y) where x and y have infinite decimal expansions x = a_1 a_2 a_3 ... [/tex] and y = b_1 b_2 b_3 ..., then you can combine these into a unique real number z = a_1 b_1 a_2 b_2 a_3 b_3 ....<br />
<br /> You need to be careful here because a real number can have more than one decimal expansion (for example, 0.5000... = 0.4999...) so the mapping as written is not well defined. I believe this problem can be avoided by always choosing the expansion that does not terminate. This of course necessitates a bit of work to show that every real number has exactly one nonterminating decimal expansion.
 
jbunniii said:
You need to be careful here because a real number can have more than one decimal expansion (for example, 0.5000... = 0.4999...) so the mapping as written is not well defined. I believe this problem can be avoided by always choosing the expansion that does not terminate. This of course necessitates a bit of work to show that every real number has exactly one nonterminating decimal expansion.

Great point, jbunniii, we should definitely make sure to deal with multiple expansions.

(By the way, I'm sorry that I have edited and updated my post multiple times. I'm still trying to figure out how to use the Tex features properly.)
 
Berrius said:
I have to prove that the cardinality of the set of infinite sequences of real numbers is equal to the cardinality of the set of real numbers.

You mean countably infinite, right?
 
ImaLooser said:
You mean countably infinite, right?
The real numbers are not countable.
 
jbunniii said:
The real numbers are not countable.
Sure, but the number of elements in the sequences is.

"the cardinality of the set of ([countable] infinite) sequences of real numbers"
I added "()" to clarify the structure.
 
mfb said:
Sure, but the number of elements in the sequences is.

"the cardinality of the set of ([countable] infinite) sequences of real numbers"
I added "()" to clarify the structure.
I hadn't thought to parse it that way. It makes more sense, but doesn't "infinite sequence" always mean "countably infinite sequence"?
 
  • #10
How about this:



If you accept that the Reals are uncountable and the rationals are countable, and that a
number is rational iff it has an eventually-periodic exoansion:

First show that the set , say S , of sequences in ℝN that are eventually-periodic are countable, and then set up a bijection between
N\S and the irrationals, sending a sequence to its "natural" decimal expansion.
 
  • #11
mfb said:
Sure, but the number of elements in the sequences is.

"the cardinality of the set of ([countable] infinite) sequences of real numbers"
I added "()" to clarify the structure.

By definition, a sequence is a function whose domain is a countable totally ordered set.
 
  • #12
pwsnafu said:
By definition, a sequence is a function whose domain is a countable totally ordered set.

Mathematicians do occasionally discuss "transfinite sequences" or sequences with an arbitrary index that may the uncountable. If the sequence is of real numbers and an uncountable number of the terms are nonzero, then the sum of the sequence necessarily diverges.
 
  • #13
hapefish said:
Mathematicians do occasionally discuss "transfinite sequences" or sequences with an arbitrary index that may the uncountable. If the sequence is of real numbers and an uncountable number of the terms are nonzero, then the sum of the sequence necessarily diverges.

The proper concept for that is to use the theory of nets.

Edit: just because "transfinite sequence" has the word sequence in it, does not mean it is in fact a "sequence" in the technical sense.
 
Last edited:

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 21 ·
Replies
21
Views
3K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K