Mixing
Defining uncountably infinite set
$begingroup$
Sets can be divided in to 3 types: finite, countably infinite, uncountable. All these are mutually disjoint and the definitions are as follows:
Finite set: Set with finite number of elements,
Countably infinite set: Infinite set which has the existence of bijection with the set of natural numbers,
Unountable: Infinite set which has no existence of bijection with the set of natural numbers,
Whether it is valid if I define uncountable set as follows:
Unountable: Infinite set which has existence of bijection with the set of Real numbers.
If not, what is the counter example for the above definition?
elementary-set-theory definition
asked Jan 26 at 12:08
hanugmhanugm
878621
$endgroup$
add a comment |
$begingroup$
Sets can be divided in to 3 types: finite, countably infinite, uncountable. All these are mutually disjoint and the definitions are as follows:
Finite set: Set with finite number of elements,
Countably infinite set: Infinite set which has the existence of bijection with the set of natural numbers,
Unountable: Infinite set which has no existence of bijection with the set of natural numbers,
Whether it is valid if I define uncountable set as follows:
Unountable: Infinite set which has existence of bijection with the set of Real numbers.
If not, what is the counter example for the above definition?
elementary-set-theory definition
asked Jan 26 at 12:08
hanugmhanugm
878621
$endgroup$
add a comment |
$begingroup$
Sets can be divided in to 3 types: finite, countably infinite, uncountable. All these are mutually disjoint and the definitions are as follows:
Finite set: Set with finite number of elements,
Countably infinite set: Infinite set which has the existence of bijection with the set of natural numbers,
Unountable: Infinite set which has no existence of bijection with the set of natural numbers,
Whether it is valid if I define uncountable set as follows:
Unountable: Infinite set which has existence of bijection with the set of Real numbers.
If not, what is the counter example for the above definition?
elementary-set-theory definition
asked Jan 26 at 12:08
hanugmhanugm
878621
$endgroup$
Sets can be divided in to 3 types: finite, countably infinite, uncountable. All these are mutually disjoint and the definitions are as follows:
Finite set: Set with finite number of elements,
Countably infinite set: Infinite set which has the existence of bijection with the set of natural numbers,
Unountable: Infinite set which has no existence of bijection with the set of natural numbers,
Whether it is valid if I define uncountable set as follows:
Unountable: Infinite set which has existence of bijection with the set of Real numbers.
If not, what is the counter example for the above definition?
elementary-set-theory definition
elementary-set-theory definition
asked Jan 26 at 12:08
hanugmhanugm
878621
asked Jan 26 at 12:08
hanugmhanugm
878621
asked Jan 26 at 12:08
hanugmhanugm
878621
asked Jan 26 at 12:08
hanugmhanugm
878621
asked Jan 26 at 12:08
hanugmhanugm
878621
878621
add a comment |
add a comment |
2 Answers
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Expressed purely in words: where this goes wrong is that there are infinitely many "sizes" (cardinalities) of uncountable set.
That's because any set has too many subsets to pair each one with an element of the set. So for any set there's a set with a bigger cardinality, namely the set of its subsets.
(Note that this is even true of the empty set, which has one subset and no element to pair it with.)
The set of sets of real numbers is therefore too big to be paired with the real numbers, and is your counterexample.
answered Jan 26 at 12:46
timtfjtimtfj
2,478420
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It is not valid. You can look at the set of the parts of the real numbers $$mathcal{P}(mathbb{R})={Avert Asubseteqmathbb{R}}.$$ It is a fairly standard exercice to show that for any set $X$, you never have a bijection $Xtomathcal{P}(X)$.
answered Jan 26 at 12:11
J.FJ.F
36713
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
Expressed purely in words: where this goes wrong is that there are infinitely many "sizes" (cardinalities) of uncountable set.
That's because any set has too many subsets to pair each one with an element of the set. So for any set there's a set with a bigger cardinality, namely the set of its subsets.
(Note that this is even true of the empty set, which has one subset and no element to pair it with.)
The set of sets of real numbers is therefore too big to be paired with the real numbers, and is your counterexample.
answered Jan 26 at 12:46
timtfjtimtfj
2,478420
$endgroup$
add a comment |
$begingroup$
Expressed purely in words: where this goes wrong is that there are infinitely many "sizes" (cardinalities) of uncountable set.
That's because any set has too many subsets to pair each one with an element of the set. So for any set there's a set with a bigger cardinality, namely the set of its subsets.
(Note that this is even true of the empty set, which has one subset and no element to pair it with.)
The set of sets of real numbers is therefore too big to be paired with the real numbers, and is your counterexample.
answered Jan 26 at 12:46
timtfjtimtfj
2,478420
$endgroup$
add a comment |
$begingroup$
Expressed purely in words: where this goes wrong is that there are infinitely many "sizes" (cardinalities) of uncountable set.
That's because any set has too many subsets to pair each one with an element of the set. So for any set there's a set with a bigger cardinality, namely the set of its subsets.
(Note that this is even true of the empty set, which has one subset and no element to pair it with.)
The set of sets of real numbers is therefore too big to be paired with the real numbers, and is your counterexample.
answered Jan 26 at 12:46
timtfjtimtfj
2,478420
$endgroup$
Expressed purely in words: where this goes wrong is that there are infinitely many "sizes" (cardinalities) of uncountable set.
That's because any set has too many subsets to pair each one with an element of the set. So for any set there's a set with a bigger cardinality, namely the set of its subsets.
(Note that this is even true of the empty set, which has one subset and no element to pair it with.)
The set of sets of real numbers is therefore too big to be paired with the real numbers, and is your counterexample.
answered Jan 26 at 12:46
timtfjtimtfj
2,478420
answered Jan 26 at 12:46
timtfjtimtfj
2,478420
answered Jan 26 at 12:46
timtfjtimtfj
2,478420
answered Jan 26 at 12:46
timtfjtimtfj
2,478420
2,478420
add a comment |
add a comment |
$begingroup$
It is not valid. You can look at the set of the parts of the real numbers $$mathcal{P}(mathbb{R})={Avert Asubseteqmathbb{R}}.$$ It is a fairly standard exercice to show that for any set $X$, you never have a bijection $Xtomathcal{P}(X)$.
answered Jan 26 at 12:11
J.FJ.F
36713
$endgroup$
add a comment |
$begingroup$
It is not valid. You can look at the set of the parts of the real numbers $$mathcal{P}(mathbb{R})={Avert Asubseteqmathbb{R}}.$$ It is a fairly standard exercice to show that for any set $X$, you never have a bijection $Xtomathcal{P}(X)$.
answered Jan 26 at 12:11
J.FJ.F
36713
$endgroup$
add a comment |
$begingroup$
It is not valid. You can look at the set of the parts of the real numbers $$mathcal{P}(mathbb{R})={Avert Asubseteqmathbb{R}}.$$ It is a fairly standard exercice to show that for any set $X$, you never have a bijection $Xtomathcal{P}(X)$.
answered Jan 26 at 12:11
J.FJ.F
36713
$endgroup$
It is not valid. You can look at the set of the parts of the real numbers $$mathcal{P}(mathbb{R})={Avert Asubseteqmathbb{R}}.$$ It is a fairly standard exercice to show that for any set $X$, you never have a bijection $Xtomathcal{P}(X)$.
answered Jan 26 at 12:11
J.FJ.F
36713
answered Jan 26 at 12:11
J.FJ.F
36713
answered Jan 26 at 12:11
J.FJ.F
36713
answered Jan 26 at 12:11
J.FJ.F
36713
36713
add a comment |
add a comment |
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