Proving that $A$ is countably infinite from another statement

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I am trying to learn Real Analysis from lectures available online: Mathematics-Real Analysis(nptel)



The professor tries to prove that (3) implies (1) in the following (in his 4th Lecture of 1st module):



  1. $A$ is countably infinite,

  2. $exists$ a subset $B$ of $mathbbN$ and a map $f:B to A$ that is onto,

  3. $exists$ a subset $C$ of $mathbbN$ and a map $g:A to C$ that is one-one.

Proof: (At 22:00 of the above video)



Consider the map $Ato g(A)$



Now, this is onto and hence, $Aapprox g(A)$



also, it is clear that $g(A)subseteq mathbbN$ and we know that every subset of $mathbbN$ is countable hence, g(A) is countable and particularly countably infinite



therefore, $A$ is countably infinite



My Doubt: why $g(A)$ is countably infinite ?




real-analysis proof-explanation




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  • 1




    Perhaps $A$ is already assumed to be infinite. Failing that, the implication is false.
    – lulu
    Jul 21 at 12:58














up vote
0
down vote

favorite












I am trying to learn Real Analysis from lectures available online: Mathematics-Real Analysis(nptel)



The professor tries to prove that (3) implies (1) in the following (in his 4th Lecture of 1st module):



  1. $A$ is countably infinite,

  2. $exists$ a subset $B$ of $mathbbN$ and a map $f:B to A$ that is onto,

  3. $exists$ a subset $C$ of $mathbbN$ and a map $g:A to C$ that is one-one.

Proof: (At 22:00 of the above video)



Consider the map $Ato g(A)$



Now, this is onto and hence, $Aapprox g(A)$



also, it is clear that $g(A)subseteq mathbbN$ and we know that every subset of $mathbbN$ is countable hence, g(A) is countable and particularly countably infinite



therefore, $A$ is countably infinite



My Doubt: why $g(A)$ is countably infinite ?




real-analysis proof-explanation




share|cite|improve this question















  • 1




    Perhaps $A$ is already assumed to be infinite. Failing that, the implication is false.
    – lulu
    Jul 21 at 12:58












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am trying to learn Real Analysis from lectures available online: Mathematics-Real Analysis(nptel)



The professor tries to prove that (3) implies (1) in the following (in his 4th Lecture of 1st module):



  1. $A$ is countably infinite,

  2. $exists$ a subset $B$ of $mathbbN$ and a map $f:B to A$ that is onto,

  3. $exists$ a subset $C$ of $mathbbN$ and a map $g:A to C$ that is one-one.

Proof: (At 22:00 of the above video)



Consider the map $Ato g(A)$



Now, this is onto and hence, $Aapprox g(A)$



also, it is clear that $g(A)subseteq mathbbN$ and we know that every subset of $mathbbN$ is countable hence, g(A) is countable and particularly countably infinite



therefore, $A$ is countably infinite



My Doubt: why $g(A)$ is countably infinite ?




real-analysis proof-explanation




share|cite|improve this question











I am trying to learn Real Analysis from lectures available online: Mathematics-Real Analysis(nptel)



The professor tries to prove that (3) implies (1) in the following (in his 4th Lecture of 1st module):



  1. $A$ is countably infinite,

  2. $exists$ a subset $B$ of $mathbbN$ and a map $f:B to A$ that is onto,

  3. $exists$ a subset $C$ of $mathbbN$ and a map $g:A to C$ that is one-one.

Proof: (At 22:00 of the above video)



Consider the map $Ato g(A)$



Now, this is onto and hence, $Aapprox g(A)$



also, it is clear that $g(A)subseteq mathbbN$ and we know that every subset of $mathbbN$ is countable hence, g(A) is countable and particularly countably infinite



therefore, $A$ is countably infinite



My Doubt: why $g(A)$ is countably infinite ?





real-analysis proof-explanation





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 21 at 12:28









Mrigank Shekhar Pathak

50219




50219







  • 1




    Perhaps $A$ is already assumed to be infinite. Failing that, the implication is false.
    – lulu
    Jul 21 at 12:58












  • 1




    Perhaps $A$ is already assumed to be infinite. Failing that, the implication is false.
    – lulu
    Jul 21 at 12:58







1




1




Perhaps $A$ is already assumed to be infinite. Failing that, the implication is false.
– lulu
Jul 21 at 12:58




Perhaps $A$ is already assumed to be infinite. Failing that, the implication is false.
– lulu
Jul 21 at 12:58










1 Answer
1






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0
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3) implies 1) is false. $C$ and $A$ could both have one element each in which case there is a one-to one map from $A$ to $C$.






share|cite|improve this answer





















  • have you seen the video I have mentioned, because in that the professor seems to claim this; might be I have interpreted something incorrectly, so please have a look at it
    – Mrigank Shekhar Pathak
    Jul 21 at 13:06










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1 Answer
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active

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oldest

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up vote
0
down vote













3) implies 1) is false. $C$ and $A$ could both have one element each in which case there is a one-to one map from $A$ to $C$.






share|cite|improve this answer





















  • have you seen the video I have mentioned, because in that the professor seems to claim this; might be I have interpreted something incorrectly, so please have a look at it
    – Mrigank Shekhar Pathak
    Jul 21 at 13:06














up vote
0
down vote













3) implies 1) is false. $C$ and $A$ could both have one element each in which case there is a one-to one map from $A$ to $C$.






share|cite|improve this answer





















  • have you seen the video I have mentioned, because in that the professor seems to claim this; might be I have interpreted something incorrectly, so please have a look at it
    – Mrigank Shekhar Pathak
    Jul 21 at 13:06












up vote
0
down vote










up vote
0
down vote









3) implies 1) is false. $C$ and $A$ could both have one element each in which case there is a one-to one map from $A$ to $C$.






share|cite|improve this answer













3) implies 1) is false. $C$ and $A$ could both have one element each in which case there is a one-to one map from $A$ to $C$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 21 at 12:41









Kavi Rama Murthy

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20.6k2830











  • have you seen the video I have mentioned, because in that the professor seems to claim this; might be I have interpreted something incorrectly, so please have a look at it
    – Mrigank Shekhar Pathak
    Jul 21 at 13:06
















  • have you seen the video I have mentioned, because in that the professor seems to claim this; might be I have interpreted something incorrectly, so please have a look at it
    – Mrigank Shekhar Pathak
    Jul 21 at 13:06















have you seen the video I have mentioned, because in that the professor seems to claim this; might be I have interpreted something incorrectly, so please have a look at it
– Mrigank Shekhar Pathak
Jul 21 at 13:06




have you seen the video I have mentioned, because in that the professor seems to claim this; might be I have interpreted something incorrectly, so please have a look at it
– Mrigank Shekhar Pathak
Jul 21 at 13:06












 

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