All open intervals are equivalent with respect to cardinality means that $mathbbR$ is equivalent to any open interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Comparing infinite sets (of real numbers)If T is an infinite subset of $mathbbN$ show that there is a 1-1 mapping of T onto $mathbbN$Proof that a subset of R has same cardinality as RBaire Category Theorem: What should we really prove there?How to prove that $mathbbQ$ is not the intersection of a countable collection of open sets.Is there a bijection from a bounded open interval of $mathbbQ$ onto $mathbbQ$?Does the class of all finite unions of closed-open intervals on $mathbbR$ form a ring sets?How do I prove that for any set $A$, $|A| < |mathbbN|$ implies that $A$ is finite?Countability of any set with cardinality larger than that of $mathbb N$Prove that the set of all monotone functions on $[0,1]$ has same cardinality as $mathbb R$
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All open intervals are equivalent with respect to cardinality means that $mathbbR$ is equivalent to any open interval
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Comparing infinite sets (of real numbers)If T is an infinite subset of $mathbbN$ show that there is a 1-1 mapping of T onto $mathbbN$Proof that a subset of R has same cardinality as RBaire Category Theorem: What should we really prove there?How to prove that $mathbbQ$ is not the intersection of a countable collection of open sets.Is there a bijection from a bounded open interval of $mathbbQ$ onto $mathbbQ$?Does the class of all finite unions of closed-open intervals on $mathbbR$ form a ring sets?How do I prove that for any set $A$, $|A| < |mathbbN|$ implies that $A$ is finite?Countability of any set with cardinality larger than that of $mathbb N$Prove that the set of all monotone functions on $[0,1]$ has same cardinality as $mathbb R$
$begingroup$
I am reading a chapter on elementary set theory and it says:
"Any open interval $(a,b)$ is equivalent to any other open interval $(c,d)$ by the function (...). Thus $mathbbR$ as a set is equivalent to any open interval of real numbers."
If I find a $f$ bijective s.t. $f: (a,b) rightarrow mathbbR$ for some $(a,b)$, it is trivial.
But can I infer the assumption immediately, without finding such an $f$ ? I don't know how to bring in the "infinite" characteristic of $mathbbR$ to the bounded intervals.
elementary-set-theory
edited Mar 26 at 16:09
gt6989b
36k22557
asked Mar 26 at 16:04
QuantaurixQuantaurix
285
$endgroup$
add a comment |
$begingroup$
I am reading a chapter on elementary set theory and it says:
"Any open interval $(a,b)$ is equivalent to any other open interval $(c,d)$ by the function (...). Thus $mathbbR$ as a set is equivalent to any open interval of real numbers."
If I find a $f$ bijective s.t. $f: (a,b) rightarrow mathbbR$ for some $(a,b)$, it is trivial.
But can I infer the assumption immediately, without finding such an $f$ ? I don't know how to bring in the "infinite" characteristic of $mathbbR$ to the bounded intervals.
elementary-set-theory
edited Mar 26 at 16:09
gt6989b
36k22557
asked Mar 26 at 16:04
QuantaurixQuantaurix
285
$endgroup$
$begingroup$
You can explicitly show a bijection between $mathbbR$ and a bounded interval. For example $tan(x)$ maps $(-pi/2,pi/2)$ to $mathbbR$ bijectively. If you only knew that all bounded intervals are equinumerous and would like to show indirectly that the implies that they have the same cardinality as $mathbbR$ you could do so by arguing that $(0,1)$ has the same cardinality as $(-n,n)$ for all $ninmathbbN$ and that $mathbbR=bigcup_n=0^infty(-n,n)$, to conclude from there that $mathbbR$ has cardinality not more than any $(-n,n)$.
$endgroup$
– user647486
Mar 26 at 16:11
add a comment |
$begingroup$
I am reading a chapter on elementary set theory and it says:
"Any open interval $(a,b)$ is equivalent to any other open interval $(c,d)$ by the function (...). Thus $mathbbR$ as a set is equivalent to any open interval of real numbers."
If I find a $f$ bijective s.t. $f: (a,b) rightarrow mathbbR$ for some $(a,b)$, it is trivial.
But can I infer the assumption immediately, without finding such an $f$ ? I don't know how to bring in the "infinite" characteristic of $mathbbR$ to the bounded intervals.
elementary-set-theory
edited Mar 26 at 16:09
gt6989b
36k22557
asked Mar 26 at 16:04
QuantaurixQuantaurix
285
$endgroup$
I am reading a chapter on elementary set theory and it says:
"Any open interval $(a,b)$ is equivalent to any other open interval $(c,d)$ by the function (...). Thus $mathbbR$ as a set is equivalent to any open interval of real numbers."
If I find a $f$ bijective s.t. $f: (a,b) rightarrow mathbbR$ for some $(a,b)$, it is trivial.
But can I infer the assumption immediately, without finding such an $f$ ? I don't know how to bring in the "infinite" characteristic of $mathbbR$ to the bounded intervals.
elementary-set-theory
elementary-set-theory
edited Mar 26 at 16:09
gt6989b
36k22557
asked Mar 26 at 16:04
QuantaurixQuantaurix
285
edited Mar 26 at 16:09
gt6989b
36k22557
asked Mar 26 at 16:04
QuantaurixQuantaurix
285
edited Mar 26 at 16:09
gt6989b
36k22557
edited Mar 26 at 16:09
gt6989b
36k22557
edited Mar 26 at 16:09
gt6989b
36k22557
36k22557
asked Mar 26 at 16:04
QuantaurixQuantaurix
285
asked Mar 26 at 16:04
QuantaurixQuantaurix
285
asked Mar 26 at 16:04
QuantaurixQuantaurix
285
285
$begingroup$
You can explicitly show a bijection between $mathbbR$ and a bounded interval. For example $tan(x)$ maps $(-pi/2,pi/2)$ to $mathbbR$ bijectively. If you only knew that all bounded intervals are equinumerous and would like to show indirectly that the implies that they have the same cardinality as $mathbbR$ you could do so by arguing that $(0,1)$ has the same cardinality as $(-n,n)$ for all $ninmathbbN$ and that $mathbbR=bigcup_n=0^infty(-n,n)$, to conclude from there that $mathbbR$ has cardinality not more than any $(-n,n)$.
$endgroup$
– user647486
Mar 26 at 16:11
add a comment |
$begingroup$
You can explicitly show a bijection between $mathbbR$ and a bounded interval. For example $tan(x)$ maps $(-pi/2,pi/2)$ to $mathbbR$ bijectively. If you only knew that all bounded intervals are equinumerous and would like to show indirectly that the implies that they have the same cardinality as $mathbbR$ you could do so by arguing that $(0,1)$ has the same cardinality as $(-n,n)$ for all $ninmathbbN$ and that $mathbbR=bigcup_n=0^infty(-n,n)$, to conclude from there that $mathbbR$ has cardinality not more than any $(-n,n)$.
$endgroup$
– user647486
Mar 26 at 16:11
$begingroup$
You can explicitly show a bijection between $mathbbR$ and a bounded interval. For example $tan(x)$ maps $(-pi/2,pi/2)$ to $mathbbR$ bijectively. If you only knew that all bounded intervals are equinumerous and would like to show indirectly that the implies that they have the same cardinality as $mathbbR$ you could do so by arguing that $(0,1)$ has the same cardinality as $(-n,n)$ for all $ninmathbbN$ and that $mathbbR=bigcup_n=0^infty(-n,n)$, to conclude from there that $mathbbR$ has cardinality not more than any $(-n,n)$.
$endgroup$
– user647486
Mar 26 at 16:11
$begingroup$
You can explicitly show a bijection between $mathbbR$ and a bounded interval. For example $tan(x)$ maps $(-pi/2,pi/2)$ to $mathbbR$ bijectively. If you only knew that all bounded intervals are equinumerous and would like to show indirectly that the implies that they have the same cardinality as $mathbbR$ you could do so by arguing that $(0,1)$ has the same cardinality as $(-n,n)$ for all $ninmathbbN$ and that $mathbbR=bigcup_n=0^infty(-n,n)$, to conclude from there that $mathbbR$ has cardinality not more than any $(-n,n)$.
$endgroup$
– user647486
Mar 26 at 16:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It seems to me that you are asking how to prove that there is a bijection between $mathbb R$ and $(a,b)$ without actually defining such a functions. I don't see a way of doing that. On the other hand, $arctan$ is a bijection between $mathbb R$ and $left(-fracpi2,fracpi2right)$. And it is easy to find a bijection between $left(-fracpi2,fracpi2right)$ and any interval $(a,b)$, and so…
answered Mar 26 at 16:10
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
$begingroup$
Yes, I thought (maybe) there's a way to construct such a bijection between $mathbbR$ and $(a,b)$ which "arises from" a certainly possible bijection between $(a,b)$ and some $(c,d)$, e.g. by stretching $(c,d)$ to $mathbbR$ (rather vague idea). Problem is, ofc, that one cannot translate fixed c,d boundaries to infinity in such a function
$endgroup$
– Quantaurix
Mar 26 at 16:15
add a comment |
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1 Answer
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$begingroup$
It seems to me that you are asking how to prove that there is a bijection between $mathbb R$ and $(a,b)$ without actually defining such a functions. I don't see a way of doing that. On the other hand, $arctan$ is a bijection between $mathbb R$ and $left(-fracpi2,fracpi2right)$. And it is easy to find a bijection between $left(-fracpi2,fracpi2right)$ and any interval $(a,b)$, and so…
answered Mar 26 at 16:10
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
$begingroup$
Yes, I thought (maybe) there's a way to construct such a bijection between $mathbbR$ and $(a,b)$ which "arises from" a certainly possible bijection between $(a,b)$ and some $(c,d)$, e.g. by stretching $(c,d)$ to $mathbbR$ (rather vague idea). Problem is, ofc, that one cannot translate fixed c,d boundaries to infinity in such a function
$endgroup$
– Quantaurix
Mar 26 at 16:15
add a comment |
$begingroup$
It seems to me that you are asking how to prove that there is a bijection between $mathbb R$ and $(a,b)$ without actually defining such a functions. I don't see a way of doing that. On the other hand, $arctan$ is a bijection between $mathbb R$ and $left(-fracpi2,fracpi2right)$. And it is easy to find a bijection between $left(-fracpi2,fracpi2right)$ and any interval $(a,b)$, and so…
answered Mar 26 at 16:10
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
$begingroup$
Yes, I thought (maybe) there's a way to construct such a bijection between $mathbbR$ and $(a,b)$ which "arises from" a certainly possible bijection between $(a,b)$ and some $(c,d)$, e.g. by stretching $(c,d)$ to $mathbbR$ (rather vague idea). Problem is, ofc, that one cannot translate fixed c,d boundaries to infinity in such a function
$endgroup$
– Quantaurix
Mar 26 at 16:15
add a comment |
$begingroup$
It seems to me that you are asking how to prove that there is a bijection between $mathbb R$ and $(a,b)$ without actually defining such a functions. I don't see a way of doing that. On the other hand, $arctan$ is a bijection between $mathbb R$ and $left(-fracpi2,fracpi2right)$. And it is easy to find a bijection between $left(-fracpi2,fracpi2right)$ and any interval $(a,b)$, and so…
answered Mar 26 at 16:10
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
It seems to me that you are asking how to prove that there is a bijection between $mathbb R$ and $(a,b)$ without actually defining such a functions. I don't see a way of doing that. On the other hand, $arctan$ is a bijection between $mathbb R$ and $left(-fracpi2,fracpi2right)$. And it is easy to find a bijection between $left(-fracpi2,fracpi2right)$ and any interval $(a,b)$, and so…
answered Mar 26 at 16:10
José Carlos SantosJosé Carlos Santos
175k24134243
answered Mar 26 at 16:10
José Carlos SantosJosé Carlos Santos
175k24134243
answered Mar 26 at 16:10
José Carlos SantosJosé Carlos Santos
175k24134243
answered Mar 26 at 16:10
José Carlos SantosJosé Carlos Santos
175k24134243
175k24134243
$begingroup$
Yes, I thought (maybe) there's a way to construct such a bijection between $mathbbR$ and $(a,b)$ which "arises from" a certainly possible bijection between $(a,b)$ and some $(c,d)$, e.g. by stretching $(c,d)$ to $mathbbR$ (rather vague idea). Problem is, ofc, that one cannot translate fixed c,d boundaries to infinity in such a function
$endgroup$
– Quantaurix
Mar 26 at 16:15
add a comment |
$begingroup$
Yes, I thought (maybe) there's a way to construct such a bijection between $mathbbR$ and $(a,b)$ which "arises from" a certainly possible bijection between $(a,b)$ and some $(c,d)$, e.g. by stretching $(c,d)$ to $mathbbR$ (rather vague idea). Problem is, ofc, that one cannot translate fixed c,d boundaries to infinity in such a function
$endgroup$
– Quantaurix
Mar 26 at 16:15
$begingroup$
Yes, I thought (maybe) there's a way to construct such a bijection between $mathbbR$ and $(a,b)$ which "arises from" a certainly possible bijection between $(a,b)$ and some $(c,d)$, e.g. by stretching $(c,d)$ to $mathbbR$ (rather vague idea). Problem is, ofc, that one cannot translate fixed c,d boundaries to infinity in such a function
$endgroup$
– Quantaurix
Mar 26 at 16:15
$begingroup$
Yes, I thought (maybe) there's a way to construct such a bijection between $mathbbR$ and $(a,b)$ which "arises from" a certainly possible bijection between $(a,b)$ and some $(c,d)$, e.g. by stretching $(c,d)$ to $mathbbR$ (rather vague idea). Problem is, ofc, that one cannot translate fixed c,d boundaries to infinity in such a function
$endgroup$
– Quantaurix
Mar 26 at 16:15
add a comment |
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You can explicitly show a bijection between $mathbbR$ and a bounded interval. For example $tan(x)$ maps $(-pi/2,pi/2)$ to $mathbbR$ bijectively. If you only knew that all bounded intervals are equinumerous and would like to show indirectly that the implies that they have the same cardinality as $mathbbR$ you could do so by arguing that $(0,1)$ has the same cardinality as $(-n,n)$ for all $ninmathbbN$ and that $mathbbR=bigcup_n=0^infty(-n,n)$, to conclude from there that $mathbbR$ has cardinality not more than any $(-n,n)$.
$endgroup$
– user647486
Mar 26 at 16:11