Countable Infinite and Uncountable Infinite sets
$begingroup$
Mark each statement as TRUE, FALSE, or UNKNOWN
(a) $|Bbb{R}| < aleph_1$
(b) $|Bbb{R}| = aleph_1$
(c) $|P(Bbb{R})| > aleph_1$
Could someone explain to me the reasoning based on whatever the answer is for each one because I do not fully understand Countable Infinite and Uncountable Infinite sets
infinite-groups
asked Dec 19 '18 at 6:31
ViseromViserom
123
$endgroup$
add a comment |
$begingroup$
Mark each statement as TRUE, FALSE, or UNKNOWN
(a) $|Bbb{R}| < aleph_1$
(b) $|Bbb{R}| = aleph_1$
(c) $|P(Bbb{R})| > aleph_1$
Could someone explain to me the reasoning based on whatever the answer is for each one because I do not fully understand Countable Infinite and Uncountable Infinite sets
infinite-groups
asked Dec 19 '18 at 6:31
ViseromViserom
123
$endgroup$
$begingroup$
Then read up on countable infinite vs. uncountable infinite. There is nothing we can say that can't be said better in a good text.
$endgroup$
– fleablood
Dec 19 '18 at 6:37
$begingroup$
@fleablood have read about it and even watched youtube clips, but still don't know how to answer the questions given.
$endgroup$
– Viserom
Dec 19 '18 at 6:41
$begingroup$
Then it's up to you to ask us a question we can answer for you. We can't read your mind and know why you don't understand it.
$endgroup$
– fleablood
Dec 19 '18 at 6:45
add a comment |
$begingroup$
Mark each statement as TRUE, FALSE, or UNKNOWN
(a) $|Bbb{R}| < aleph_1$
(b) $|Bbb{R}| = aleph_1$
(c) $|P(Bbb{R})| > aleph_1$
Could someone explain to me the reasoning based on whatever the answer is for each one because I do not fully understand Countable Infinite and Uncountable Infinite sets
infinite-groups
asked Dec 19 '18 at 6:31
ViseromViserom
123
$endgroup$
Mark each statement as TRUE, FALSE, or UNKNOWN
(a) $|Bbb{R}| < aleph_1$
(b) $|Bbb{R}| = aleph_1$
(c) $|P(Bbb{R})| > aleph_1$
Could someone explain to me the reasoning based on whatever the answer is for each one because I do not fully understand Countable Infinite and Uncountable Infinite sets
infinite-groups
infinite-groups
asked Dec 19 '18 at 6:31
ViseromViserom
123
asked Dec 19 '18 at 6:31
ViseromViserom
123
asked Dec 19 '18 at 6:31
ViseromViserom
123
asked Dec 19 '18 at 6:31
ViseromViserom
123
asked Dec 19 '18 at 6:31
ViseromViserom
123
123
$begingroup$
Then read up on countable infinite vs. uncountable infinite. There is nothing we can say that can't be said better in a good text.
$endgroup$
– fleablood
Dec 19 '18 at 6:37
$begingroup$
@fleablood have read about it and even watched youtube clips, but still don't know how to answer the questions given.
$endgroup$
– Viserom
Dec 19 '18 at 6:41
$begingroup$
Then it's up to you to ask us a question we can answer for you. We can't read your mind and know why you don't understand it.
$endgroup$
– fleablood
Dec 19 '18 at 6:45
add a comment |
$begingroup$
Then read up on countable infinite vs. uncountable infinite. There is nothing we can say that can't be said better in a good text.
$endgroup$
– fleablood
Dec 19 '18 at 6:37
$begingroup$
@fleablood have read about it and even watched youtube clips, but still don't know how to answer the questions given.
$endgroup$
– Viserom
Dec 19 '18 at 6:41
$begingroup$
Then it's up to you to ask us a question we can answer for you. We can't read your mind and know why you don't understand it.
$endgroup$
– fleablood
Dec 19 '18 at 6:45
$begingroup$
Then read up on countable infinite vs. uncountable infinite. There is nothing we can say that can't be said better in a good text.
$endgroup$
– fleablood
Dec 19 '18 at 6:37
$begingroup$
Then read up on countable infinite vs. uncountable infinite. There is nothing we can say that can't be said better in a good text.
$endgroup$
– fleablood
Dec 19 '18 at 6:37
$begingroup$
@fleablood have read about it and even watched youtube clips, but still don't know how to answer the questions given.
$endgroup$
– Viserom
Dec 19 '18 at 6:41
$begingroup$
@fleablood have read about it and even watched youtube clips, but still don't know how to answer the questions given.
$endgroup$
– Viserom
Dec 19 '18 at 6:41
$begingroup$
Then it's up to you to ask us a question we can answer for you. We can't read your mind and know why you don't understand it.
$endgroup$
– fleablood
Dec 19 '18 at 6:45
$begingroup$
Then it's up to you to ask us a question we can answer for you. We can't read your mind and know why you don't understand it.
$endgroup$
– fleablood
Dec 19 '18 at 6:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$aleph_1$ by definition is the first uncountable cardinal number.
So $|A| < aleph_1$ by definition means that $A$ is a countable set.
The reals are not countable.
We (should) know that $|mathbb{R}| = |P(mathbb{N})| = 2^{aleph_0}> aleph_0$. So in particular $2^{aleph_0} ge aleph_1$.
The so-called continuum hypothesis (which is independent of ZFC, so "unknown" in your list) states that $2^{aleph_0} = aleph_1$.
By Cantor's theorem $|P(mathbb{R})| = 2^{|mathbb{R}|}= 2^{2^{aleph_0}} > 2^{aleph_0} ge aleph_1$.
answered Dec 19 '18 at 7:28
Henno BrandsmaHenno Brandsma
114k348124
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add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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votes
$begingroup$
$aleph_1$ by definition is the first uncountable cardinal number.
So $|A| < aleph_1$ by definition means that $A$ is a countable set.
The reals are not countable.
We (should) know that $|mathbb{R}| = |P(mathbb{N})| = 2^{aleph_0}> aleph_0$. So in particular $2^{aleph_0} ge aleph_1$.
The so-called continuum hypothesis (which is independent of ZFC, so "unknown" in your list) states that $2^{aleph_0} = aleph_1$.
By Cantor's theorem $|P(mathbb{R})| = 2^{|mathbb{R}|}= 2^{2^{aleph_0}} > 2^{aleph_0} ge aleph_1$.
answered Dec 19 '18 at 7:28
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
add a comment |
$begingroup$
$aleph_1$ by definition is the first uncountable cardinal number.
So $|A| < aleph_1$ by definition means that $A$ is a countable set.
The reals are not countable.
We (should) know that $|mathbb{R}| = |P(mathbb{N})| = 2^{aleph_0}> aleph_0$. So in particular $2^{aleph_0} ge aleph_1$.
The so-called continuum hypothesis (which is independent of ZFC, so "unknown" in your list) states that $2^{aleph_0} = aleph_1$.
By Cantor's theorem $|P(mathbb{R})| = 2^{|mathbb{R}|}= 2^{2^{aleph_0}} > 2^{aleph_0} ge aleph_1$.
answered Dec 19 '18 at 7:28
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
add a comment |
$begingroup$
$aleph_1$ by definition is the first uncountable cardinal number.
So $|A| < aleph_1$ by definition means that $A$ is a countable set.
The reals are not countable.
We (should) know that $|mathbb{R}| = |P(mathbb{N})| = 2^{aleph_0}> aleph_0$. So in particular $2^{aleph_0} ge aleph_1$.
The so-called continuum hypothesis (which is independent of ZFC, so "unknown" in your list) states that $2^{aleph_0} = aleph_1$.
By Cantor's theorem $|P(mathbb{R})| = 2^{|mathbb{R}|}= 2^{2^{aleph_0}} > 2^{aleph_0} ge aleph_1$.
answered Dec 19 '18 at 7:28
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
$aleph_1$ by definition is the first uncountable cardinal number.
So $|A| < aleph_1$ by definition means that $A$ is a countable set.
The reals are not countable.
We (should) know that $|mathbb{R}| = |P(mathbb{N})| = 2^{aleph_0}> aleph_0$. So in particular $2^{aleph_0} ge aleph_1$.
The so-called continuum hypothesis (which is independent of ZFC, so "unknown" in your list) states that $2^{aleph_0} = aleph_1$.
By Cantor's theorem $|P(mathbb{R})| = 2^{|mathbb{R}|}= 2^{2^{aleph_0}} > 2^{aleph_0} ge aleph_1$.
answered Dec 19 '18 at 7:28
Henno BrandsmaHenno Brandsma
114k348124
answered Dec 19 '18 at 7:28
Henno BrandsmaHenno Brandsma
114k348124
answered Dec 19 '18 at 7:28
Henno BrandsmaHenno Brandsma
114k348124
answered Dec 19 '18 at 7:28
Henno BrandsmaHenno Brandsma
114k348124
114k348124
add a comment |
add a comment |
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$begingroup$
Then read up on countable infinite vs. uncountable infinite. There is nothing we can say that can't be said better in a good text.
$endgroup$
– fleablood
Dec 19 '18 at 6:37
$begingroup$
@fleablood have read about it and even watched youtube clips, but still don't know how to answer the questions given.
$endgroup$
– Viserom
Dec 19 '18 at 6:41
$begingroup$
Then it's up to you to ask us a question we can answer for you. We can't read your mind and know why you don't understand it.
$endgroup$
– fleablood
Dec 19 '18 at 6:45