Mixing
Apostol proof for $mathbb{Q}$ being countable.
$begingroup$
I am trying to understand a proof from Mathematical Analysis by Apostol for the following theorem:
The set of rationals $mathbb{Q}$ is countable.
Here is the proof (I rewrote a few things):
Let $A_n$ be the set of positive rational numbers that have denominator $n$. Then the following is true: $$mathbb{Q}^+=bigcup_{i=1}^infty A_i$$ Since each $A_i$ is countable, $mathbb{Q}^+$ is countable. Similarly for $mathbb{Q}^-$ and ${0}$. Then $mathbb{Q}=mathbb{Q}^-cup{0}cupmathbb{Q}^+$ is countable.
For some reason, I cannot understand the bold part. How is that each $A_{i}$ is countable?
real-analysis elementary-set-theory proof-explanation rational-numbers
edited Jan 7 at 15:19
Andrés E. Caicedo
65.2k8158247
asked Jan 7 at 3:46
numericalorangenumericalorange
1,735311
$endgroup$
add a comment |
$begingroup$
I am trying to understand a proof from Mathematical Analysis by Apostol for the following theorem:
The set of rationals $mathbb{Q}$ is countable.
Here is the proof (I rewrote a few things):
Let $A_n$ be the set of positive rational numbers that have denominator $n$. Then the following is true: $$mathbb{Q}^+=bigcup_{i=1}^infty A_i$$ Since each $A_i$ is countable, $mathbb{Q}^+$ is countable. Similarly for $mathbb{Q}^-$ and ${0}$. Then $mathbb{Q}=mathbb{Q}^-cup{0}cupmathbb{Q}^+$ is countable.
For some reason, I cannot understand the bold part. How is that each $A_{i}$ is countable?
real-analysis elementary-set-theory proof-explanation rational-numbers
edited Jan 7 at 15:19
Andrés E. Caicedo
65.2k8158247
asked Jan 7 at 3:46
numericalorangenumericalorange
1,735311
$endgroup$
$begingroup$
Think about A_2 for concreteness. There is one element of A_2 for every positive integer that you put in the denominator (or zero, if you don't want to count non-reduced fractions, but it doesn't matter either way). Explicitly, the elements of A_2 are 1/2, 2/2, 3/2, 4/2, 5/2, and so on
$endgroup$
– Brennan Vincent
Jan 7 at 4:07
add a comment |
$begingroup$
I am trying to understand a proof from Mathematical Analysis by Apostol for the following theorem:
The set of rationals $mathbb{Q}$ is countable.
Here is the proof (I rewrote a few things):
Let $A_n$ be the set of positive rational numbers that have denominator $n$. Then the following is true: $$mathbb{Q}^+=bigcup_{i=1}^infty A_i$$ Since each $A_i$ is countable, $mathbb{Q}^+$ is countable. Similarly for $mathbb{Q}^-$ and ${0}$. Then $mathbb{Q}=mathbb{Q}^-cup{0}cupmathbb{Q}^+$ is countable.
For some reason, I cannot understand the bold part. How is that each $A_{i}$ is countable?
real-analysis elementary-set-theory proof-explanation rational-numbers
edited Jan 7 at 15:19
Andrés E. Caicedo
65.2k8158247
asked Jan 7 at 3:46
numericalorangenumericalorange
1,735311
$endgroup$
I am trying to understand a proof from Mathematical Analysis by Apostol for the following theorem:
The set of rationals $mathbb{Q}$ is countable.
Here is the proof (I rewrote a few things):
Let $A_n$ be the set of positive rational numbers that have denominator $n$. Then the following is true: $$mathbb{Q}^+=bigcup_{i=1}^infty A_i$$ Since each $A_i$ is countable, $mathbb{Q}^+$ is countable. Similarly for $mathbb{Q}^-$ and ${0}$. Then $mathbb{Q}=mathbb{Q}^-cup{0}cupmathbb{Q}^+$ is countable.
For some reason, I cannot understand the bold part. How is that each $A_{i}$ is countable?
real-analysis elementary-set-theory proof-explanation rational-numbers
real-analysis elementary-set-theory proof-explanation rational-numbers
edited Jan 7 at 15:19
Andrés E. Caicedo
65.2k8158247
asked Jan 7 at 3:46
numericalorangenumericalorange
1,735311
edited Jan 7 at 15:19
Andrés E. Caicedo
65.2k8158247
asked Jan 7 at 3:46
numericalorangenumericalorange
1,735311
edited Jan 7 at 15:19
Andrés E. Caicedo
65.2k8158247
edited Jan 7 at 15:19
Andrés E. Caicedo
65.2k8158247
edited Jan 7 at 15:19
Andrés E. Caicedo
65.2k8158247
65.2k8158247
asked Jan 7 at 3:46
numericalorangenumericalorange
1,735311
asked Jan 7 at 3:46
numericalorangenumericalorange
1,735311
asked Jan 7 at 3:46
numericalorangenumericalorange
1,735311
1,735311
$begingroup$
Think about A_2 for concreteness. There is one element of A_2 for every positive integer that you put in the denominator (or zero, if you don't want to count non-reduced fractions, but it doesn't matter either way). Explicitly, the elements of A_2 are 1/2, 2/2, 3/2, 4/2, 5/2, and so on
$endgroup$
– Brennan Vincent
Jan 7 at 4:07
add a comment |
$begingroup$
Think about A_2 for concreteness. There is one element of A_2 for every positive integer that you put in the denominator (or zero, if you don't want to count non-reduced fractions, but it doesn't matter either way). Explicitly, the elements of A_2 are 1/2, 2/2, 3/2, 4/2, 5/2, and so on
$endgroup$
– Brennan Vincent
Jan 7 at 4:07
$begingroup$
Think about A_2 for concreteness. There is one element of A_2 for every positive integer that you put in the denominator (or zero, if you don't want to count non-reduced fractions, but it doesn't matter either way). Explicitly, the elements of A_2 are 1/2, 2/2, 3/2, 4/2, 5/2, and so on
$endgroup$
– Brennan Vincent
Jan 7 at 4:07
$begingroup$
Think about A_2 for concreteness. There is one element of A_2 for every positive integer that you put in the denominator (or zero, if you don't want to count non-reduced fractions, but it doesn't matter either way). Explicitly, the elements of A_2 are 1/2, 2/2, 3/2, 4/2, 5/2, and so on
$endgroup$
– Brennan Vincent
Jan 7 at 4:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider the map $f_i:A_irightarrow mathbb{N}$ defined by $f({aover i})=a$, it is injective. This implies that $A_i$ has a cardinality of a subset of $mathbb{N}$ so it is countable.
answered Jan 7 at 3:49
Tsemo AristideTsemo Aristide
57.3k11444
$endgroup$
add a comment |
$begingroup$
Rephrasing:
Fix $i$ , $i in mathbb{Z^+}$.
$A_i= ${$1/i,2/i,3/i,.......$}.
$f(mathbb{Z^+}) rightarrow A_i,$
$f(n)= n/i$, $i$ fixed, is a bijection.
$A_i$ is countable.
answered Jan 7 at 6:52
Peter SzilasPeter Szilas
11.1k2821
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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$begingroup$
Consider the map $f_i:A_irightarrow mathbb{N}$ defined by $f({aover i})=a$, it is injective. This implies that $A_i$ has a cardinality of a subset of $mathbb{N}$ so it is countable.
answered Jan 7 at 3:49
Tsemo AristideTsemo Aristide
57.3k11444
$endgroup$
add a comment |
$begingroup$
Consider the map $f_i:A_irightarrow mathbb{N}$ defined by $f({aover i})=a$, it is injective. This implies that $A_i$ has a cardinality of a subset of $mathbb{N}$ so it is countable.
answered Jan 7 at 3:49
Tsemo AristideTsemo Aristide
57.3k11444
$endgroup$
add a comment |
$begingroup$
Consider the map $f_i:A_irightarrow mathbb{N}$ defined by $f({aover i})=a$, it is injective. This implies that $A_i$ has a cardinality of a subset of $mathbb{N}$ so it is countable.
answered Jan 7 at 3:49
Tsemo AristideTsemo Aristide
57.3k11444
$endgroup$
Consider the map $f_i:A_irightarrow mathbb{N}$ defined by $f({aover i})=a$, it is injective. This implies that $A_i$ has a cardinality of a subset of $mathbb{N}$ so it is countable.
answered Jan 7 at 3:49
Tsemo AristideTsemo Aristide
57.3k11444
answered Jan 7 at 3:49
Tsemo AristideTsemo Aristide
57.3k11444
answered Jan 7 at 3:49
Tsemo AristideTsemo Aristide
57.3k11444
answered Jan 7 at 3:49
Tsemo AristideTsemo Aristide
57.3k11444
57.3k11444
add a comment |
add a comment |
$begingroup$
Rephrasing:
Fix $i$ , $i in mathbb{Z^+}$.
$A_i= ${$1/i,2/i,3/i,.......$}.
$f(mathbb{Z^+}) rightarrow A_i,$
$f(n)= n/i$, $i$ fixed, is a bijection.
$A_i$ is countable.
answered Jan 7 at 6:52
Peter SzilasPeter Szilas
11.1k2821
$endgroup$
add a comment |
$begingroup$
Rephrasing:
Fix $i$ , $i in mathbb{Z^+}$.
$A_i= ${$1/i,2/i,3/i,.......$}.
$f(mathbb{Z^+}) rightarrow A_i,$
$f(n)= n/i$, $i$ fixed, is a bijection.
$A_i$ is countable.
answered Jan 7 at 6:52
Peter SzilasPeter Szilas
11.1k2821
$endgroup$
add a comment |
$begingroup$
Rephrasing:
Fix $i$ , $i in mathbb{Z^+}$.
$A_i= ${$1/i,2/i,3/i,.......$}.
$f(mathbb{Z^+}) rightarrow A_i,$
$f(n)= n/i$, $i$ fixed, is a bijection.
$A_i$ is countable.
answered Jan 7 at 6:52
Peter SzilasPeter Szilas
11.1k2821
$endgroup$
Rephrasing:
Fix $i$ , $i in mathbb{Z^+}$.
$A_i= ${$1/i,2/i,3/i,.......$}.
$f(mathbb{Z^+}) rightarrow A_i,$
$f(n)= n/i$, $i$ fixed, is a bijection.
$A_i$ is countable.
answered Jan 7 at 6:52
Peter SzilasPeter Szilas
11.1k2821
edited Jan 7 at 7:19
answered Jan 7 at 6:52
Peter SzilasPeter Szilas
11.1k2821
answered Jan 7 at 6:52
Peter SzilasPeter Szilas
11.1k2821
answered Jan 7 at 6:52
Peter SzilasPeter Szilas
11.1k2821
11.1k2821
add a comment |
add a comment |
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$begingroup$
Think about A_2 for concreteness. There is one element of A_2 for every positive integer that you put in the denominator (or zero, if you don't want to count non-reduced fractions, but it doesn't matter either way). Explicitly, the elements of A_2 are 1/2, 2/2, 3/2, 4/2, 5/2, and so on
$endgroup$
– Brennan Vincent
Jan 7 at 4:07