Fdtxjr

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ssTTsSTtRrriinInnnnNNNIiinngg

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Prove that $|mathbbNtimes mathbbR| = |mathbbR|$

Can there exist an injective function from $mathbb R$ to $(0,1)$?The cartesian product $mathbbN times mathbbN$ is countableCompare the cardinalities of $2^(2^mathbb N)$ and $(2^mathbb N)^mathbb N$$mathbbN times mathbbN$ is countable?Number of relations and number of functions on an infinite setConstruct an injective function $f:[a,b]times[c,d]rightarrowmathbbR$ for some $a,b,c,dinmathbbR$The usual bijection between $[0,1]$ and $[0,1]times[0,1]$Why could there be no injection between $mathbbZtimesmathbbZ$ and $mathbbZ$, and between $mathbbZ*mathbbZ$ and $mathbbZ$?How did Royden prove the surjectiveness of the function mapping $mathbb Ntimes mathbb N$ and $mathbb Q$ to $mathbb N$?Prove that the space of functions on [0,1] with |f(x)| $leq$ 1 has cardinality 2$^aleph$Are there uncountably many injective functions from $mathbb N$ to $mathbb N$?

1

$begingroup$

I'm trying to prove the equality between the cardinalities:
$|mathbbNtimes mathbbR| = |mathbbR|$ ($mathbbNtimes mathbbR$ is cartesian product from $mathbbN$ to $mathbbR$).

I tried doing that by building injective functions first from $mathbbNtimes mathbbR$ to $(0,1)$ and then from $(0,1)$ to $mathbbNtimes mathbbR$ so:

$F: mathbbNtimes mathbbR rightarrow (0,1)$ is given by :
$$F(1 , 0 .abcd....) mapsto 0.abcd..$$
$$F(2 , 0.abcde....) mapsto 0.abcde..$$
for example
($0.abcd..$ represents a real number in $(0,1)$

I'm not sure if that's right..

elementary-set-theory cardinals

share|cite|improve this question

edited Mar 21 at 9:00

Andrews

1,2812422

asked Mar 21 at 8:00

yoav amenouyoav amenou

224

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've added MathJax to your post for you, which I hope makes it easier to read. It would be good if you could check it over and make sure I've not changed anything though
  $endgroup$
  – postmortes
  Mar 21 at 8:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is definitely something that was asked before. Admittedly, though, it's not something that's easy to search for.
  $endgroup$
  – Asaf Karagila♦
  Mar 21 at 8:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The function, $F$, you've constructed is not injective since (by your own example) $F(1,0.abcde...)=F(2,0.abcde...)$. Try instead letting $F:mathbb Ntimes mathbb Rrightarrow mathbb R$ rather than just $F:mathbb Ntimes mathbb Rrightarrow (0,1)$
  $endgroup$
  – Cardioid_Ass_22
  Mar 21 at 8:53
  

  
  
  
  

add a comment | 

1

$begingroup$

I'm trying to prove the equality between the cardinalities:
$|mathbbNtimes mathbbR| = |mathbbR|$ ($mathbbNtimes mathbbR$ is cartesian product from $mathbbN$ to $mathbbR$).

I tried doing that by building injective functions first from $mathbbNtimes mathbbR$ to $(0,1)$ and then from $(0,1)$ to $mathbbNtimes mathbbR$ so:

$F: mathbbNtimes mathbbR rightarrow (0,1)$ is given by :
$$F(1 , 0 .abcd....) mapsto 0.abcd..$$
$$F(2 , 0.abcde....) mapsto 0.abcde..$$
for example
($0.abcd..$ represents a real number in $(0,1)$

I'm not sure if that's right..

elementary-set-theory cardinals

share|cite|improve this question

edited Mar 21 at 9:00

Andrews

1,2812422

asked Mar 21 at 8:00

yoav amenouyoav amenou

224

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've added MathJax to your post for you, which I hope makes it easier to read. It would be good if you could check it over and make sure I've not changed anything though
  $endgroup$
  – postmortes
  Mar 21 at 8:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is definitely something that was asked before. Admittedly, though, it's not something that's easy to search for.
  $endgroup$
  – Asaf Karagila♦
  Mar 21 at 8:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The function, $F$, you've constructed is not injective since (by your own example) $F(1,0.abcde...)=F(2,0.abcde...)$. Try instead letting $F:mathbb Ntimes mathbb Rrightarrow mathbb R$ rather than just $F:mathbb Ntimes mathbb Rrightarrow (0,1)$
  $endgroup$
  – Cardioid_Ass_22
  Mar 21 at 8:53
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to prove the equality between the cardinalities:
$|mathbbNtimes mathbbR| = |mathbbR|$ ($mathbbNtimes mathbbR$ is cartesian product from $mathbbN$ to $mathbbR$).

I tried doing that by building injective functions first from $mathbbNtimes mathbbR$ to $(0,1)$ and then from $(0,1)$ to $mathbbNtimes mathbbR$ so:

$F: mathbbNtimes mathbbR rightarrow (0,1)$ is given by :
$$F(1 , 0 .abcd....) mapsto 0.abcd..$$
$$F(2 , 0.abcde....) mapsto 0.abcde..$$
for example
($0.abcd..$ represents a real number in $(0,1)$

I'm not sure if that's right..

elementary-set-theory cardinals

share|cite|improve this question

edited Mar 21 at 9:00

Andrews

1,2812422

asked Mar 21 at 8:00

yoav amenouyoav amenou

224

$endgroup$

share|cite|improve this question

edited Mar 21 at 9:00

Andrews

1,2812422

asked Mar 21 at 8:00

yoav amenouyoav amenou

224

share|cite|improve this question

edited Mar 21 at 9:00

Andrews

1,2812422

asked Mar 21 at 8:00

yoav amenouyoav amenou

224

edited Mar 21 at 9:00

Andrews

1,2812422

edited Mar 21 at 9:00

Andrews

1,2812422

asked Mar 21 at 8:00

yoav amenouyoav amenou

224

asked Mar 21 at 8:00

yoav amenouyoav amenou

224

2 Answers
2

active

oldest

votes

2

$begingroup$

(I will assume $BbbN$ starts at $1$). Use the fact that $|BbbR| = mathcalP(BbbN)$; it suffices to show that $|mathcalP(BbbN)| = |BbbN times mathcalP(BbbN)|$.

Clearly,
$$|BbbN times mathcalP(BbbN)| ge |1 times mathcalP(BbbN)| = |mathcalP(BbbN)|,$$
so it suffices to find a surjection from $mathcalP(BbbN)$ onto $BbbN times mathcalP(BbbN)$.

Define such a function as follows. Suppose $S subseteq BbbN$. If $S = emptyset$, map it to $(1, emptyset)$. Otherwise, $S$ has a minimum element. Map $S$ to $(min S, (S setminus min S) - min S)$.

We now show that this map is surjective. If we have a pair $(n, T) in BbbN times mathcalP(BbbN)$, then let $S = n cup (T + n)$. Since every $t in T$ is at least $1$, we have that $n$ is the minimum element of $S$, and clearly the surjection will map $S$ to $T$.

(It's worth pointing out that the map is a bijection too, if you restrict it to $mathcalP(BbbN) setminus emptyset $.)

share|cite|improve this answer

answered Mar 21 at 9:16

Theo BenditTheo Bendit

20.2k12353

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks! but just one thing i couldn't understand is why you added the -minS at the end (minS, (smins) [-mins]
  $endgroup$
  – yoav amenou
  Mar 21 at 10:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Otherwise it's not surjective. There would be no way to achieve, for example, $(2, 1)$ -- have a think about why that is.
  $endgroup$
  – Theo Bendit
  Mar 21 at 10:38
  

  
  
  
  

add a comment | 

1

$begingroup$

To begin with, try solving something simpler: find an injective function $f:mathbb Rtomathbb (0,1)$ (this is not trivial, but is commonly known; see, e.g., This MSE thread, or using some manipulation of the $arctan$ function).

Once you have this, you can `cook up' a new function $F:mathbb Ntimesmathbb Rto mathbb R$ by setting
$$F(n,x)=f(x)+ntext for any ninmathbb Ntext and xin mathbb R.$$
It should be a simple check that the new function $F$ is injective.

share|cite|improve this answer

answered Mar 21 at 9:18

kneidellkneidell

1,7631321

$endgroup$

add a comment | 

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2

$begingroup$

(I will assume $BbbN$ starts at $1$). Use the fact that $|BbbR| = mathcalP(BbbN)$; it suffices to show that $|mathcalP(BbbN)| = |BbbN times mathcalP(BbbN)|$.

Clearly,
$$|BbbN times mathcalP(BbbN)| ge |1 times mathcalP(BbbN)| = |mathcalP(BbbN)|,$$
so it suffices to find a surjection from $mathcalP(BbbN)$ onto $BbbN times mathcalP(BbbN)$.

Define such a function as follows. Suppose $S subseteq BbbN$. If $S = emptyset$, map it to $(1, emptyset)$. Otherwise, $S$ has a minimum element. Map $S$ to $(min S, (S setminus min S) - min S)$.

We now show that this map is surjective. If we have a pair $(n, T) in BbbN times mathcalP(BbbN)$, then let $S = n cup (T + n)$. Since every $t in T$ is at least $1$, we have that $n$ is the minimum element of $S$, and clearly the surjection will map $S$ to $T$.

(It's worth pointing out that the map is a bijection too, if you restrict it to $mathcalP(BbbN) setminus emptyset $.)

share|cite|improve this answer

answered Mar 21 at 9:16

Theo BenditTheo Bendit

20.2k12353

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks! but just one thing i couldn't understand is why you added the -minS at the end (minS, (smins) [-mins]
  $endgroup$
  – yoav amenou
  Mar 21 at 10:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Otherwise it's not surjective. There would be no way to achieve, for example, $(2, 1)$ -- have a think about why that is.
  $endgroup$
  – Theo Bendit
  Mar 21 at 10:38
  

  
  
  
  

add a comment | 

2

$begingroup$

(I will assume $BbbN$ starts at $1$). Use the fact that $|BbbR| = mathcalP(BbbN)$; it suffices to show that $|mathcalP(BbbN)| = |BbbN times mathcalP(BbbN)|$.

Clearly,
$$|BbbN times mathcalP(BbbN)| ge |1 times mathcalP(BbbN)| = |mathcalP(BbbN)|,$$
so it suffices to find a surjection from $mathcalP(BbbN)$ onto $BbbN times mathcalP(BbbN)$.

Define such a function as follows. Suppose $S subseteq BbbN$. If $S = emptyset$, map it to $(1, emptyset)$. Otherwise, $S$ has a minimum element. Map $S$ to $(min S, (S setminus min S) - min S)$.

We now show that this map is surjective. If we have a pair $(n, T) in BbbN times mathcalP(BbbN)$, then let $S = n cup (T + n)$. Since every $t in T$ is at least $1$, we have that $n$ is the minimum element of $S$, and clearly the surjection will map $S$ to $T$.

(It's worth pointing out that the map is a bijection too, if you restrict it to $mathcalP(BbbN) setminus emptyset $.)

share|cite|improve this answer

answered Mar 21 at 9:16

Theo BenditTheo Bendit

20.2k12353

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks! but just one thing i couldn't understand is why you added the -minS at the end (minS, (smins) [-mins]
  $endgroup$
  – yoav amenou
  Mar 21 at 10:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Otherwise it's not surjective. There would be no way to achieve, for example, $(2, 1)$ -- have a think about why that is.
  $endgroup$
  – Theo Bendit
  Mar 21 at 10:38
  

  
  
  
  

add a comment | 

$begingroup$

(I will assume $BbbN$ starts at $1$). Use the fact that $|BbbR| = mathcalP(BbbN)$; it suffices to show that $|mathcalP(BbbN)| = |BbbN times mathcalP(BbbN)|$.

Clearly,
$$|BbbN times mathcalP(BbbN)| ge |1 times mathcalP(BbbN)| = |mathcalP(BbbN)|,$$
so it suffices to find a surjection from $mathcalP(BbbN)$ onto $BbbN times mathcalP(BbbN)$.

Define such a function as follows. Suppose $S subseteq BbbN$. If $S = emptyset$, map it to $(1, emptyset)$. Otherwise, $S$ has a minimum element. Map $S$ to $(min S, (S setminus min S) - min S)$.

We now show that this map is surjective. If we have a pair $(n, T) in BbbN times mathcalP(BbbN)$, then let $S = n cup (T + n)$. Since every $t in T$ is at least $1$, we have that $n$ is the minimum element of $S$, and clearly the surjection will map $S$ to $T$.

(It's worth pointing out that the map is a bijection too, if you restrict it to $mathcalP(BbbN) setminus emptyset $.)

share|cite|improve this answer

answered Mar 21 at 9:16

Theo BenditTheo Bendit

20.2k12353

$endgroup$

share|cite|improve this answer

answered Mar 21 at 9:16

Theo BenditTheo Bendit

20.2k12353

answered Mar 21 at 9:16

Theo BenditTheo Bendit

20.2k12353

answered Mar 21 at 9:16

Theo BenditTheo Bendit

20.2k12353

1

$begingroup$

To begin with, try solving something simpler: find an injective function $f:mathbb Rtomathbb (0,1)$ (this is not trivial, but is commonly known; see, e.g., This MSE thread, or using some manipulation of the $arctan$ function).

Once you have this, you can `cook up' a new function $F:mathbb Ntimesmathbb Rto mathbb R$ by setting
$$F(n,x)=f(x)+ntext for any ninmathbb Ntext and xin mathbb R.$$
It should be a simple check that the new function $F$ is injective.

share|cite|improve this answer

answered Mar 21 at 9:18

kneidellkneidell

1,7631321

$endgroup$

add a comment | 

1

$begingroup$

To begin with, try solving something simpler: find an injective function $f:mathbb Rtomathbb (0,1)$ (this is not trivial, but is commonly known; see, e.g., This MSE thread, or using some manipulation of the $arctan$ function).

Once you have this, you can `cook up' a new function $F:mathbb Ntimesmathbb Rto mathbb R$ by setting
$$F(n,x)=f(x)+ntext for any ninmathbb Ntext and xin mathbb R.$$
It should be a simple check that the new function $F$ is injective.

share|cite|improve this answer

answered Mar 21 at 9:18

kneidellkneidell

1,7631321

$endgroup$

add a comment | 

$begingroup$

To begin with, try solving something simpler: find an injective function $f:mathbb Rtomathbb (0,1)$ (this is not trivial, but is commonly known; see, e.g., This MSE thread, or using some manipulation of the $arctan$ function).

Once you have this, you can `cook up' a new function $F:mathbb Ntimesmathbb Rto mathbb R$ by setting
$$F(n,x)=f(x)+ntext for any ninmathbb Ntext and xin mathbb R.$$
It should be a simple check that the new function $F$ is injective.

share|cite|improve this answer

answered Mar 21 at 9:18

kneidellkneidell

1,7631321

$endgroup$

share|cite|improve this answer

answered Mar 21 at 9:18

kneidellkneidell

1,7631321

answered Mar 21 at 9:18

kneidellkneidell

1,7631321

answered Mar 21 at 9:18

kneidellkneidell

1,7631321