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Proving that if a set A is denumerable and a set B that is finite and a subset of A, then $Asetminus B$ is denumerable

Basic Set-Theoretic Properties from HalmosReal Analysis, Folland Proposition 1.7 elementary familyProving that every set has a finite subset of ``smallest elements''.Set $A$ partitionable into denumerable sets implies injection from $mathbb{N}$ to $A$Relax Egoroff's Theorem to pointwise convergence a.e. and bounded a.e. pointwise limitSuppose $X$ is infinite and $A$ is a finite subset of $X$. Then $X$ and $X setminus A$ are equinumerousLet $X$ be uncountable and $A$ be a countable subset of $X$. Then $X$ and $X setminus A$ are equinumerousThe cardinality of every maximal subset of the independent set $mathcal{I}$ is same. (Matroid Theory)If a set $A$ is finite then $Acap B$ is a finite set.Proving a function from $mathbb{N}to Acup {x }$ is a bijection.

0

1

$begingroup$

Background:
This comes from the book: INTRODUCTION TO
MATHEMATICAL
PROOFS
Charles E. Roberts, Jr.
Indiana State University
Terre Haute, USA
A Transition to Advanced
Mathematics
Second Edition

A set $A$ is denumerable if and only if $Asim mathbb{N}$.

$Asim B$ if and only if there is a one-to-one correspondence (bijection) from $A$ to $B$.

Theorem 7.15 - If $A$ is a denumerable set and $B$ is a finite set, the $Acup B$ is a denumerable set.

Question:

Prove that if $A$ is a denumerable set and $B$ is a finite subset of $A$, then $Asetminus B$ is denumerable.

Attempted proof - Note that

$$begin{align*}
Asetminus B &= Acap B^c\
&= (Acup B)cap B^c
end{align*}$$

Borrowing from the comments below. Since $Acup B$ is denumerable by theorem 7.15, and any subset of a denumerable set is denumerable this implies that $Asetminus B$ is denumerable.

We know from theorem 7.15 that $A cup B$ is denumerable. I am just not sure how to show that the complement of the finite set $B$ with $Acup B$ is also denumerable.

real-analysis proof-verification

share|cite|improve this question

edited Mar 12 at 4:15

Snorrlaxxx

asked Mar 12 at 3:56

SnorrlaxxxSnorrlaxxx

1415

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Can't you just use the fact that subsets of denumerable sets are denumerable?
  $endgroup$
  – Don Thousand
  Mar 12 at 4:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that $A = (Asetminus B)cup B$. What happens if $Asetminus B$ is not denumerable?
  $endgroup$
  – Lucas Corrêa
  Mar 12 at 4:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Other approach is to use the Hilbert's Hotel Problem. I think that the idea works fine.
  $endgroup$
  – Lucas Corrêa
  Mar 12 at 4:08
  

  
  
  
  

add a comment | 

0

1

$begingroup$

Background:
This comes from the book: INTRODUCTION TO
MATHEMATICAL
PROOFS
Charles E. Roberts, Jr.
Indiana State University
Terre Haute, USA
A Transition to Advanced
Mathematics
Second Edition

A set $A$ is denumerable if and only if $Asim mathbb{N}$.

$Asim B$ if and only if there is a one-to-one correspondence (bijection) from $A$ to $B$.

Theorem 7.15 - If $A$ is a denumerable set and $B$ is a finite set, the $Acup B$ is a denumerable set.

Question:

Prove that if $A$ is a denumerable set and $B$ is a finite subset of $A$, then $Asetminus B$ is denumerable.

Attempted proof - Note that

$$begin{align*}
Asetminus B &= Acap B^c\
&= (Acup B)cap B^c
end{align*}$$

Borrowing from the comments below. Since $Acup B$ is denumerable by theorem 7.15, and any subset of a denumerable set is denumerable this implies that $Asetminus B$ is denumerable.

We know from theorem 7.15 that $A cup B$ is denumerable. I am just not sure how to show that the complement of the finite set $B$ with $Acup B$ is also denumerable.

real-analysis proof-verification

share|cite|improve this question

edited Mar 12 at 4:15

Snorrlaxxx

asked Mar 12 at 3:56

SnorrlaxxxSnorrlaxxx

1415

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Can't you just use the fact that subsets of denumerable sets are denumerable?
  $endgroup$
  – Don Thousand
  Mar 12 at 4:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that $A = (Asetminus B)cup B$. What happens if $Asetminus B$ is not denumerable?
  $endgroup$
  – Lucas Corrêa
  Mar 12 at 4:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Other approach is to use the Hilbert's Hotel Problem. I think that the idea works fine.
  $endgroup$
  – Lucas Corrêa
  Mar 12 at 4:08
  

  
  
  
  

add a comment | 

$begingroup$

Background:
This comes from the book: INTRODUCTION TO
MATHEMATICAL
PROOFS
Charles E. Roberts, Jr.
Indiana State University
Terre Haute, USA
A Transition to Advanced
Mathematics
Second Edition

A set $A$ is denumerable if and only if $Asim mathbb{N}$.

$Asim B$ if and only if there is a one-to-one correspondence (bijection) from $A$ to $B$.

Theorem 7.15 - If $A$ is a denumerable set and $B$ is a finite set, the $Acup B$ is a denumerable set.

Question:

Prove that if $A$ is a denumerable set and $B$ is a finite subset of $A$, then $Asetminus B$ is denumerable.

Attempted proof - Note that

$$begin{align*}
Asetminus B &= Acap B^c\
&= (Acup B)cap B^c
end{align*}$$

Borrowing from the comments below. Since $Acup B$ is denumerable by theorem 7.15, and any subset of a denumerable set is denumerable this implies that $Asetminus B$ is denumerable.

We know from theorem 7.15 that $A cup B$ is denumerable. I am just not sure how to show that the complement of the finite set $B$ with $Acup B$ is also denumerable.

real-analysis proof-verification

share|cite|improve this question

edited Mar 12 at 4:15

Snorrlaxxx

asked Mar 12 at 3:56

SnorrlaxxxSnorrlaxxx

1415

$endgroup$

share|cite|improve this question

edited Mar 12 at 4:15

Snorrlaxxx

asked Mar 12 at 3:56

SnorrlaxxxSnorrlaxxx

1415

share|cite|improve this question

edited Mar 12 at 4:15

Snorrlaxxx

asked Mar 12 at 3:56

SnorrlaxxxSnorrlaxxx

1415

asked Mar 12 at 3:56

SnorrlaxxxSnorrlaxxx

1415

asked Mar 12 at 3:56

SnorrlaxxxSnorrlaxxx

1415

2 Answers
2

active

oldest

votes

1

$begingroup$

Try this:

We have that there exists a bijection $f:A to mathbb{N}$

Then $g = f^{-1}:mathbb{N} to A$ is a bijection too

And $B = {b_{1},...,b_{k}} subset A$

So, we have that for every $n in mathbb{N}$, $g(n) = a_{n} in A$

And $g(n_{i}) = b_{i}$ for some $n_{i}$ with $ 1 leq i leq k$

And suppose that $n_{1}<n_{2}< dots <n_{k}$

We are going to create a bijection $h:mathbb{N} to$ $A$ $B$

Let $d in mathbb{N}cup{0}$, $d = n_{k} - k$

We have a subset with $d$ elements ${1,2,...,n_{k}}$ ${n_{1},...n_{k}} = {c_{1},...,c_{d}}$

Let $h(n) = g(c_{n})$ if $1 leq n leq d$, and $h(n) = g(n + k)$ if $n>d$

Then $h$ define a bijection between $mathbb{N}$ and $A$ $B$

share|cite|improve this answer

answered Mar 12 at 4:32

ZAFZAF

4607

$endgroup$

add a comment | 

2

$begingroup$

First, we observe that $A setminus B$ is not finite, otherwise $A = A setminus B cup B$ would be finite. Now, $A setminus B$ (being infinite) is either denumerable or non - denumerable. Since $A setminus B subseteq A$ and $A$ is denumerable, $A setminus B$ cannot be non - denumerable (A subset cannot contain "more" elements than its superset). Hence, the only possibility is that $A setminus B$ is denumerable.

share|cite|improve this answer

edited Mar 12 at 4:35

answered Mar 12 at 4:04

Aniruddha DeshmukhAniruddha Deshmukh

1,161419

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You mean $A$ $B$ cannot be non-denumerable?
  $endgroup$
  – J. W. Tanner
  Mar 12 at 4:11
  
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

Try this:

We have that there exists a bijection $f:A to mathbb{N}$

Then $g = f^{-1}:mathbb{N} to A$ is a bijection too

And $B = {b_{1},...,b_{k}} subset A$

So, we have that for every $n in mathbb{N}$, $g(n) = a_{n} in A$

And $g(n_{i}) = b_{i}$ for some $n_{i}$ with $ 1 leq i leq k$

And suppose that $n_{1}<n_{2}< dots <n_{k}$

We are going to create a bijection $h:mathbb{N} to$ $A$ $B$

Let $d in mathbb{N}cup{0}$, $d = n_{k} - k$

We have a subset with $d$ elements ${1,2,...,n_{k}}$ ${n_{1},...n_{k}} = {c_{1},...,c_{d}}$

Let $h(n) = g(c_{n})$ if $1 leq n leq d$, and $h(n) = g(n + k)$ if $n>d$

Then $h$ define a bijection between $mathbb{N}$ and $A$ $B$

share|cite|improve this answer

answered Mar 12 at 4:32

ZAFZAF

4607

$endgroup$

add a comment | 

1

$begingroup$

Try this:

We have that there exists a bijection $f:A to mathbb{N}$

Then $g = f^{-1}:mathbb{N} to A$ is a bijection too

And $B = {b_{1},...,b_{k}} subset A$

So, we have that for every $n in mathbb{N}$, $g(n) = a_{n} in A$

And $g(n_{i}) = b_{i}$ for some $n_{i}$ with $ 1 leq i leq k$

And suppose that $n_{1}<n_{2}< dots <n_{k}$

We are going to create a bijection $h:mathbb{N} to$ $A$ $B$

Let $d in mathbb{N}cup{0}$, $d = n_{k} - k$

We have a subset with $d$ elements ${1,2,...,n_{k}}$ ${n_{1},...n_{k}} = {c_{1},...,c_{d}}$

Let $h(n) = g(c_{n})$ if $1 leq n leq d$, and $h(n) = g(n + k)$ if $n>d$

Then $h$ define a bijection between $mathbb{N}$ and $A$ $B$

share|cite|improve this answer

answered Mar 12 at 4:32

ZAFZAF

4607

$endgroup$

add a comment | 

$begingroup$

Try this:

We have that there exists a bijection $f:A to mathbb{N}$

Then $g = f^{-1}:mathbb{N} to A$ is a bijection too

And $B = {b_{1},...,b_{k}} subset A$

So, we have that for every $n in mathbb{N}$, $g(n) = a_{n} in A$

And $g(n_{i}) = b_{i}$ for some $n_{i}$ with $ 1 leq i leq k$

And suppose that $n_{1}<n_{2}< dots <n_{k}$

We are going to create a bijection $h:mathbb{N} to$ $A$ $B$

Let $d in mathbb{N}cup{0}$, $d = n_{k} - k$

We have a subset with $d$ elements ${1,2,...,n_{k}}$ ${n_{1},...n_{k}} = {c_{1},...,c_{d}}$

Let $h(n) = g(c_{n})$ if $1 leq n leq d$, and $h(n) = g(n + k)$ if $n>d$

Then $h$ define a bijection between $mathbb{N}$ and $A$ $B$

share|cite|improve this answer

answered Mar 12 at 4:32

ZAFZAF

4607

$endgroup$

share|cite|improve this answer

answered Mar 12 at 4:32

ZAFZAF

4607

answered Mar 12 at 4:32

ZAFZAF

4607

answered Mar 12 at 4:32

ZAFZAF

4607

2

$begingroup$

First, we observe that $A setminus B$ is not finite, otherwise $A = A setminus B cup B$ would be finite. Now, $A setminus B$ (being infinite) is either denumerable or non - denumerable. Since $A setminus B subseteq A$ and $A$ is denumerable, $A setminus B$ cannot be non - denumerable (A subset cannot contain "more" elements than its superset). Hence, the only possibility is that $A setminus B$ is denumerable.

share|cite|improve this answer

edited Mar 12 at 4:35

answered Mar 12 at 4:04

Aniruddha DeshmukhAniruddha Deshmukh

1,161419

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You mean $A$ $B$ cannot be non-denumerable?
  $endgroup$
  – J. W. Tanner
  Mar 12 at 4:11
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

First, we observe that $A setminus B$ is not finite, otherwise $A = A setminus B cup B$ would be finite. Now, $A setminus B$ (being infinite) is either denumerable or non - denumerable. Since $A setminus B subseteq A$ and $A$ is denumerable, $A setminus B$ cannot be non - denumerable (A subset cannot contain "more" elements than its superset). Hence, the only possibility is that $A setminus B$ is denumerable.

share|cite|improve this answer

edited Mar 12 at 4:35

answered Mar 12 at 4:04

Aniruddha DeshmukhAniruddha Deshmukh

1,161419

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You mean $A$ $B$ cannot be non-denumerable?
  $endgroup$
  – J. W. Tanner
  Mar 12 at 4:11
  
  
  

  
  
  
  

add a comment | 

$begingroup$

First, we observe that $A setminus B$ is not finite, otherwise $A = A setminus B cup B$ would be finite. Now, $A setminus B$ (being infinite) is either denumerable or non - denumerable. Since $A setminus B subseteq A$ and $A$ is denumerable, $A setminus B$ cannot be non - denumerable (A subset cannot contain "more" elements than its superset). Hence, the only possibility is that $A setminus B$ is denumerable.

share|cite|improve this answer

edited Mar 12 at 4:35

answered Mar 12 at 4:04

Aniruddha DeshmukhAniruddha Deshmukh

1,161419

$endgroup$

share|cite|improve this answer

edited Mar 12 at 4:35

answered Mar 12 at 4:04

Aniruddha DeshmukhAniruddha Deshmukh

1,161419

answered Mar 12 at 4:04

Aniruddha DeshmukhAniruddha Deshmukh

1,161419

answered Mar 12 at 4:04

Aniruddha DeshmukhAniruddha Deshmukh

1,161419