cardinal product $mn=m$ if $nleq m$?

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If $m,n geq aleph_0$ and $mgeq n$, then which is right? $mn=m$ or $mnleq m$?
Why if the first not correct?
For examply, if the weight of a topological space $omega(X)=n$, then the product $omega(X^m)leq m$ or $omega(X)=m$?

general-topology cardinals

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edited Jul 16 at 5:08

GFauxPas

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asked Jul 16 at 4:27

Shen Chong

426

add a comment | 

up vote
1
down vote

favorite

If $m,n geq aleph_0$ and $mgeq n$, then which is right? $mn=m$ or $mnleq m$?
Why if the first not correct?
For examply, if the weight of a topological space $omega(X)=n$, then the product $omega(X^m)leq m$ or $omega(X)=m$?

general-topology cardinals

share|cite|improve this question

edited Jul 16 at 5:08

GFauxPas

4,11811129

asked Jul 16 at 4:27

Shen Chong

426

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

If $m,n geq aleph_0$ and $mgeq n$, then which is right? $mn=m$ or $mnleq m$?
Why if the first not correct?
For examply, if the weight of a topological space $omega(X)=n$, then the product $omega(X^m)leq m$ or $omega(X)=m$?

general-topology cardinals

share|cite|improve this question

edited Jul 16 at 5:08

GFauxPas

4,11811129

asked Jul 16 at 4:27

Shen Chong

426

If $m,n geq aleph_0$ and $mgeq n$, then which is right? $mn=m$ or $mnleq m$?
Why if the first not correct?
For examply, if the weight of a topological space $omega(X)=n$, then the product $omega(X^m)leq m$ or $omega(X)=m$?

general-topology cardinals

share|cite|improve this question

edited Jul 16 at 5:08

GFauxPas

4,11811129

asked Jul 16 at 4:27

Shen Chong

426

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 5:08

GFauxPas

4,11811129

edited Jul 16 at 5:08

GFauxPas

4,11811129

edited Jul 16 at 5:08

GFauxPas

4,11811129

4,11811129

asked Jul 16 at 4:27

Shen Chong

426

asked Jul 16 at 4:27

Shen Chong

426

asked Jul 16 at 4:27

Shen Chong

426

426

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

The product of two infinite cardinals is the maximum, so if $m ge n ge aleph_0$, $mn = m$.

If $w(X) = n$ (weight uses w = $w$ not omega= $omega$ as its symbol) then $w(X^m)$ is the weight of an infinite power, so this weight is indeed $mn = m$ (the standard base built from basic elements of a base of size $n$ of $X$ in a power of $m$ spaces has size $nm$ and when $m ge n$ we have that $w(X^m) = m$). In particular $w([0,1]^m) = m$ for all infinite cardinals $m$, as $w([0,1]) = aleph_0$.

share|cite|improve this answer

answered Jul 16 at 4:43

Henno Brandsma

91.6k342100

add a comment | 

up vote
3
down vote

Both are right.

If $x=y$, then $xleq y$, but also since $leq$ is antisymmetric, and clearly $mleq mn$,¹ we get that $mnleq m$ implies $mn=m$.

---

_{(1) In this case, at least, where $n$ is non-zero.}

share|cite|improve this answer

answered Jul 16 at 4:33

Asaf Karagila♦

292k31403733

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. While I see another theorem on the textbook of Engelking's $General$ $Topology$ as follows:
  – Shen Chong
  Jul 18 at 8:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. While I see another theorem on the textbook of Engelking's $General$ $Topology$ as follows: If $w(X_s)leq mgeqaleph_0$ for every $sin S$ and $|S|leq m$, then $w(Pi_sin SX_s)leq m$. Why not $=$?
  – Shen Chong
  Jul 18 at 8:22
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

The product of two infinite cardinals is the maximum, so if $m ge n ge aleph_0$, $mn = m$.

If $w(X) = n$ (weight uses w = $w$ not omega= $omega$ as its symbol) then $w(X^m)$ is the weight of an infinite power, so this weight is indeed $mn = m$ (the standard base built from basic elements of a base of size $n$ of $X$ in a power of $m$ spaces has size $nm$ and when $m ge n$ we have that $w(X^m) = m$). In particular $w([0,1]^m) = m$ for all infinite cardinals $m$, as $w([0,1]) = aleph_0$.

share|cite|improve this answer

answered Jul 16 at 4:43

Henno Brandsma

91.6k342100

add a comment | 

up vote
2
down vote



accepted

The product of two infinite cardinals is the maximum, so if $m ge n ge aleph_0$, $mn = m$.

If $w(X) = n$ (weight uses w = $w$ not omega= $omega$ as its symbol) then $w(X^m)$ is the weight of an infinite power, so this weight is indeed $mn = m$ (the standard base built from basic elements of a base of size $n$ of $X$ in a power of $m$ spaces has size $nm$ and when $m ge n$ we have that $w(X^m) = m$). In particular $w([0,1]^m) = m$ for all infinite cardinals $m$, as $w([0,1]) = aleph_0$.

share|cite|improve this answer

answered Jul 16 at 4:43

Henno Brandsma

91.6k342100

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

The product of two infinite cardinals is the maximum, so if $m ge n ge aleph_0$, $mn = m$.

If $w(X) = n$ (weight uses w = $w$ not omega= $omega$ as its symbol) then $w(X^m)$ is the weight of an infinite power, so this weight is indeed $mn = m$ (the standard base built from basic elements of a base of size $n$ of $X$ in a power of $m$ spaces has size $nm$ and when $m ge n$ we have that $w(X^m) = m$). In particular $w([0,1]^m) = m$ for all infinite cardinals $m$, as $w([0,1]) = aleph_0$.

share|cite|improve this answer

answered Jul 16 at 4:43

Henno Brandsma

91.6k342100

The product of two infinite cardinals is the maximum, so if $m ge n ge aleph_0$, $mn = m$.

If $w(X) = n$ (weight uses w = $w$ not omega= $omega$ as its symbol) then $w(X^m)$ is the weight of an infinite power, so this weight is indeed $mn = m$ (the standard base built from basic elements of a base of size $n$ of $X$ in a power of $m$ spaces has size $nm$ and when $m ge n$ we have that $w(X^m) = m$). In particular $w([0,1]^m) = m$ for all infinite cardinals $m$, as $w([0,1]) = aleph_0$.

share|cite|improve this answer

answered Jul 16 at 4:43

Henno Brandsma

91.6k342100

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 4:43

Henno Brandsma

91.6k342100

answered Jul 16 at 4:43

Henno Brandsma

91.6k342100

answered Jul 16 at 4:43

Henno Brandsma

91.6k342100

91.6k342100

add a comment | 

add a comment | 

up vote
3
down vote

Both are right.

If $x=y$, then $xleq y$, but also since $leq$ is antisymmetric, and clearly $mleq mn$,¹ we get that $mnleq m$ implies $mn=m$.

---

_{(1) In this case, at least, where $n$ is non-zero.}

share|cite|improve this answer

answered Jul 16 at 4:33

Asaf Karagila♦

292k31403733

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. While I see another theorem on the textbook of Engelking's $General$ $Topology$ as follows:
  – Shen Chong
  Jul 18 at 8:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. While I see another theorem on the textbook of Engelking's $General$ $Topology$ as follows: If $w(X_s)leq mgeqaleph_0$ for every $sin S$ and $|S|leq m$, then $w(Pi_sin SX_s)leq m$. Why not $=$?
  – Shen Chong
  Jul 18 at 8:22
  

  
  
  
  

add a comment | 

up vote
3
down vote

Both are right.

If $x=y$, then $xleq y$, but also since $leq$ is antisymmetric, and clearly $mleq mn$,¹ we get that $mnleq m$ implies $mn=m$.

---

_{(1) In this case, at least, where $n$ is non-zero.}

share|cite|improve this answer

answered Jul 16 at 4:33

Asaf Karagila♦

292k31403733

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. While I see another theorem on the textbook of Engelking's $General$ $Topology$ as follows:
  – Shen Chong
  Jul 18 at 8:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. While I see another theorem on the textbook of Engelking's $General$ $Topology$ as follows: If $w(X_s)leq mgeqaleph_0$ for every $sin S$ and $|S|leq m$, then $w(Pi_sin SX_s)leq m$. Why not $=$?
  – Shen Chong
  Jul 18 at 8:22
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Both are right.

If $x=y$, then $xleq y$, but also since $leq$ is antisymmetric, and clearly $mleq mn$,¹ we get that $mnleq m$ implies $mn=m$.

---

_{(1) In this case, at least, where $n$ is non-zero.}

share|cite|improve this answer

answered Jul 16 at 4:33

Asaf Karagila♦

292k31403733

Both are right.

If $x=y$, then $xleq y$, but also since $leq$ is antisymmetric, and clearly $mleq mn$,¹ we get that $mnleq m$ implies $mn=m$.

---

_{(1) In this case, at least, where $n$ is non-zero.}

share|cite|improve this answer

answered Jul 16 at 4:33

Asaf Karagila♦

292k31403733

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 4:33

Asaf Karagila♦

292k31403733

answered Jul 16 at 4:33

Asaf Karagila♦

292k31403733

answered Jul 16 at 4:33

Asaf Karagila♦

292k31403733

292k31403733

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. While I see another theorem on the textbook of Engelking's $General$ $Topology$ as follows:
  – Shen Chong
  Jul 18 at 8:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. While I see another theorem on the textbook of Engelking's $General$ $Topology$ as follows: If $w(X_s)leq mgeqaleph_0$ for every $sin S$ and $|S|leq m$, then $w(Pi_sin SX_s)leq m$. Why not $=$?
  – Shen Chong
  Jul 18 at 8:22
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. While I see another theorem on the textbook of Engelking's $General$ $Topology$ as follows:
  – Shen Chong
  Jul 18 at 8:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer. While I see another theorem on the textbook of Engelking's $General$ $Topology$ as follows: If $w(X_s)leq mgeqaleph_0$ for every $sin S$ and $|S|leq m$, then $w(Pi_sin SX_s)leq m$. Why not $=$?
  – Shen Chong
  Jul 18 at 8:22
  

  
  
  
  

Thanks for your answer. While I see another theorem on the textbook of Engelking's $General$ $Topology$ as follows:
– Shen Chong
Jul 18 at 8:15

Thanks for your answer. While I see another theorem on the textbook of Engelking's $General$ $Topology$ as follows:
– Shen Chong
Jul 18 at 8:15

Thanks for your answer. While I see another theorem on the textbook of Engelking's $General$ $Topology$ as follows: If $w(X_s)leq mgeqaleph_0$ for every $sin S$ and $|S|leq m$, then $w(Pi_sin SX_s)leq m$. Why not $=$?
– Shen Chong
Jul 18 at 8:22

Thanks for your answer. While I see another theorem on the textbook of Engelking's $General$ $Topology$ as follows: If $w(X_s)leq mgeqaleph_0$ for every $sin S$ and $|S|leq m$, then $w(Pi_sin SX_s)leq m$. Why not $=$?
– Shen Chong
Jul 18 at 8:22

add a comment | 

 

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