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How can we identify a set is infinite when proving Bolzano–Weierstrass theorem?
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I'm proving the Bolzano–Weierstrass theorem. It says
$$
textIf A text : bounded and infinite set, text then A text has at least one limit point.
$$
So the proof goes like the following.
Since $A subset mathbbR^N$ is bounded, it can be a subset of a box $B_1 = I_1 times cdots times I_N$, where $I_i$ is an interval in $mathbbR$. Let's divide the $B_1$ into $2^N$ sub-boxes. Then, at least one sub-box has the infinite elements of $A$. Let's say that sub-box as $B_2$. Let's say we can repeat this procedure for $B_3, B_4, ldots$
Well, so proof ends as we prove some $x$ exists in $cap_n=1^inftyB_n$ and it is a limit point of $A$.
But what I'm wondering is, how can we identify the $B_2$? (and $B_3$, and so on)
Yes, it's clear to me that such set exists, but the existence itself does not guarantee us to identify the set (that is, to select and label it as $B_2$), since we don't have any tools to discern if $B_2$ is infinite or not.
It sounds a little bit philosophical, but is there anyone to help?
real-analysis infinity
asked Jul 28 at 7:47
moreblue
1738
 |Â
show 11 more comments
up vote
7
down vote
favorite
I'm proving the Bolzano–Weierstrass theorem. It says
$$
textIf A text : bounded and infinite set, text then A text has at least one limit point.
$$
So the proof goes like the following.
Since $A subset mathbbR^N$ is bounded, it can be a subset of a box $B_1 = I_1 times cdots times I_N$, where $I_i$ is an interval in $mathbbR$. Let's divide the $B_1$ into $2^N$ sub-boxes. Then, at least one sub-box has the infinite elements of $A$. Let's say that sub-box as $B_2$. Let's say we can repeat this procedure for $B_3, B_4, ldots$
Well, so proof ends as we prove some $x$ exists in $cap_n=1^inftyB_n$ and it is a limit point of $A$.
But what I'm wondering is, how can we identify the $B_2$? (and $B_3$, and so on)
Yes, it's clear to me that such set exists, but the existence itself does not guarantee us to identify the set (that is, to select and label it as $B_2$), since we don't have any tools to discern if $B_2$ is infinite or not.
It sounds a little bit philosophical, but is there anyone to help?
real-analysis infinity
asked Jul 28 at 7:47
moreblue
1738
You say at least one sub-box has. Also there is a missing step somewhere, the highest $B_k$ mentioned is $B_2$ but you then write $xin cap_nB_n$ we don't know what any $B_k, k>2$ is at this point.
– mathreadler
Jul 28 at 7:53
3
What do you mean exactly by "identify"? And why do we need to do so?
– Anguepa
Jul 28 at 7:53
@Angeupa What I mean by "identify" is, what we need to do before "Let's say that sub-box as $B_2$". I can prove at least one such set exists, but how can we actually pick that set out and label it as $B_2$?
– moreblue
Jul 28 at 8:01
4
It is an existence proof not a construction proof. It only shows that such a point must exist, not that there must be only one or how to find it (if it is only one). The points must sum up and the only way to sum to infinity is if one or more of the subboxes sums to infinity since if none of them did then it would be finite number of points which would violate the assumption.
– mathreadler
Jul 28 at 8:04
Since there are finitely many sub-boxes ($2^N$), one can (in principle) check each box until we find one that contains infinitely many elements.
– Brahadeesh
Jul 28 at 8:04
 |Â
show 11 more comments
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I'm proving the Bolzano–Weierstrass theorem. It says
$$
textIf A text : bounded and infinite set, text then A text has at least one limit point.
$$
So the proof goes like the following.
Since $A subset mathbbR^N$ is bounded, it can be a subset of a box $B_1 = I_1 times cdots times I_N$, where $I_i$ is an interval in $mathbbR$. Let's divide the $B_1$ into $2^N$ sub-boxes. Then, at least one sub-box has the infinite elements of $A$. Let's say that sub-box as $B_2$. Let's say we can repeat this procedure for $B_3, B_4, ldots$
Well, so proof ends as we prove some $x$ exists in $cap_n=1^inftyB_n$ and it is a limit point of $A$.
But what I'm wondering is, how can we identify the $B_2$? (and $B_3$, and so on)
Yes, it's clear to me that such set exists, but the existence itself does not guarantee us to identify the set (that is, to select and label it as $B_2$), since we don't have any tools to discern if $B_2$ is infinite or not.
It sounds a little bit philosophical, but is there anyone to help?
real-analysis infinity
asked Jul 28 at 7:47
moreblue
1738
I'm proving the Bolzano–Weierstrass theorem. It says
$$
textIf A text : bounded and infinite set, text then A text has at least one limit point.
$$
So the proof goes like the following.
Since $A subset mathbbR^N$ is bounded, it can be a subset of a box $B_1 = I_1 times cdots times I_N$, where $I_i$ is an interval in $mathbbR$. Let's divide the $B_1$ into $2^N$ sub-boxes. Then, at least one sub-box has the infinite elements of $A$. Let's say that sub-box as $B_2$. Let's say we can repeat this procedure for $B_3, B_4, ldots$
Well, so proof ends as we prove some $x$ exists in $cap_n=1^inftyB_n$ and it is a limit point of $A$.
But what I'm wondering is, how can we identify the $B_2$? (and $B_3$, and so on)
Yes, it's clear to me that such set exists, but the existence itself does not guarantee us to identify the set (that is, to select and label it as $B_2$), since we don't have any tools to discern if $B_2$ is infinite or not.
It sounds a little bit philosophical, but is there anyone to help?
real-analysis infinity
asked Jul 28 at 7:47
moreblue
1738
edited Jul 28 at 8:35
asked Jul 28 at 7:47
moreblue
1738
asked Jul 28 at 7:47
moreblue
1738
asked Jul 28 at 7:47
moreblue
1738
1738
You say at least one sub-box has. Also there is a missing step somewhere, the highest $B_k$ mentioned is $B_2$ but you then write $xin cap_nB_n$ we don't know what any $B_k, k>2$ is at this point.
– mathreadler
Jul 28 at 7:53
3
What do you mean exactly by "identify"? And why do we need to do so?
– Anguepa
Jul 28 at 7:53
@Angeupa What I mean by "identify" is, what we need to do before "Let's say that sub-box as $B_2$". I can prove at least one such set exists, but how can we actually pick that set out and label it as $B_2$?
– moreblue
Jul 28 at 8:01
4
It is an existence proof not a construction proof. It only shows that such a point must exist, not that there must be only one or how to find it (if it is only one). The points must sum up and the only way to sum to infinity is if one or more of the subboxes sums to infinity since if none of them did then it would be finite number of points which would violate the assumption.
– mathreadler
Jul 28 at 8:04
Since there are finitely many sub-boxes ($2^N$), one can (in principle) check each box until we find one that contains infinitely many elements.
– Brahadeesh
Jul 28 at 8:04
 |Â
show 11 more comments
You say at least one sub-box has. Also there is a missing step somewhere, the highest $B_k$ mentioned is $B_2$ but you then write $xin cap_nB_n$ we don't know what any $B_k, k>2$ is at this point.
– mathreadler
Jul 28 at 7:53
3
What do you mean exactly by "identify"? And why do we need to do so?
– Anguepa
Jul 28 at 7:53
@Angeupa What I mean by "identify" is, what we need to do before "Let's say that sub-box as $B_2$". I can prove at least one such set exists, but how can we actually pick that set out and label it as $B_2$?
– moreblue
Jul 28 at 8:01
4
It is an existence proof not a construction proof. It only shows that such a point must exist, not that there must be only one or how to find it (if it is only one). The points must sum up and the only way to sum to infinity is if one or more of the subboxes sums to infinity since if none of them did then it would be finite number of points which would violate the assumption.
– mathreadler
Jul 28 at 8:04
Since there are finitely many sub-boxes ($2^N$), one can (in principle) check each box until we find one that contains infinitely many elements.
– Brahadeesh
Jul 28 at 8:04
You say at least one sub-box has. Also there is a missing step somewhere, the highest $B_k$ mentioned is $B_2$ but you then write $xin cap_nB_n$ we don't know what any $B_k, k>2$ is at this point.
– mathreadler
Jul 28 at 7:53
You say at least one sub-box has. Also there is a missing step somewhere, the highest $B_k$ mentioned is $B_2$ but you then write $xin cap_nB_n$ we don't know what any $B_k, k>2$ is at this point.
– mathreadler
Jul 28 at 7:53
3
3
What do you mean exactly by "identify"? And why do we need to do so?
– Anguepa
Jul 28 at 7:53
What do you mean exactly by "identify"? And why do we need to do so?
– Anguepa
Jul 28 at 7:53
@Angeupa What I mean by "identify" is, what we need to do before "Let's say that sub-box as $B_2$". I can prove at least one such set exists, but how can we actually pick that set out and label it as $B_2$?
– moreblue
Jul 28 at 8:01
@Angeupa What I mean by "identify" is, what we need to do before "Let's say that sub-box as $B_2$". I can prove at least one such set exists, but how can we actually pick that set out and label it as $B_2$?
– moreblue
Jul 28 at 8:01
4
4
It is an existence proof not a construction proof. It only shows that such a point must exist, not that there must be only one or how to find it (if it is only one). The points must sum up and the only way to sum to infinity is if one or more of the subboxes sums to infinity since if none of them did then it would be finite number of points which would violate the assumption.
– mathreadler
Jul 28 at 8:04
It is an existence proof not a construction proof. It only shows that such a point must exist, not that there must be only one or how to find it (if it is only one). The points must sum up and the only way to sum to infinity is if one or more of the subboxes sums to infinity since if none of them did then it would be finite number of points which would violate the assumption.
– mathreadler
Jul 28 at 8:04
Since there are finitely many sub-boxes ($2^N$), one can (in principle) check each box until we find one that contains infinitely many elements.
– Brahadeesh
Jul 28 at 8:04
Since there are finitely many sub-boxes ($2^N$), one can (in principle) check each box until we find one that contains infinitely many elements.
– Brahadeesh
Jul 28 at 8:04
 |Â
show 11 more comments
3 Answers
3
active
oldest
votes
up vote
7
down vote
accepted
It is not surprising that there is no constructive way to identify, at each stage, which of the $2^n$ boxes holds infinitely many points—because, from the outset, the premiss of the theorem assumes nothing about where the points are: we know only that there are infinitely many of them. There are thus at least continuum-many possibilities for how the points might be distributed. All we can do is make the logical observation that, at every stage, the boxes cannot all hold a finite number of points, and therefore at least one of them holds an infinite number. If we need to identify just which such a box might be, we might be able to do so if we were given a precise construction of $A$; but we are given no such information.
answered Jul 28 at 8:45
John Bentin
10.8k22350
add a comment |Â
up vote
7
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It seems that you understand why at least one box contains infinitely many elements of $A$, and your difficulty is only in finding which box that is (or finding one such box if there are several of them). That is, it's a question of existence versus finding. Notice, though, that the theorem you're proving says only that there exists a limit point. It doesn't claim anything about finding a limit point.
To prove the existence of a limit point, it's enough to have the existence of an appropriate box at every stage of the process.
If you wanted to prove a stronger theorem saying that one can find a limit point, then you'd have to worry about finding an appropriate box at every stage. Any alleged theorem of that sort would have to be preceded by information about (1) exactly how the set $A$ is presented and (2) exactly what it means to "find" a point. So such a theorem would not only be harder to prove but considerably more complicated to state than the Bolzano-Weierstrass theorem.
answered Jul 28 at 12:51
Andreas Blass
47.5k348104
add a comment |Â
up vote
4
down vote
It is in fact a philosophical question. The "paradoxes of infinity" have puzzled people since ancient times. In 1851 Bolzano himself has written a book with the title "Paradoxien des Unendlichen" (in English "Paradoxes of the Infinite").
In your proof you start with an infinite set $A$. But how can you know that is infinite? Is it explicitly given by listing its elements? In broad terms let us understand by "listing" any rule that allows you to decide whether a given element $x$ belongs to $A$ or not.
If you have such a listing, then you will be able to identify $B_2, B_3, ...$. If not, you are not even able to identify $A$. In that case it is only possible to argue on an abstract level: If you put infinitely many things into finitely many boxes, than at least one box must contain infinitely many things. Imagine a "person" has put the things into the boxes, but you didn't watch the process. Then you will not know which box contains infinitely many things, but you will know that at least one box must contain infinitely many things. If you don't like to consider "infinitely many things", you can also consider $10$ things in $2$ boxes: You know that at least one must contain $5$ things or more, but you don't know which box. It is similar to the shell game: You know that the ball is placed beneath one the shells, but would you bet beneath which?
answered Jul 28 at 8:55
Paul Frost
3,603420
Regarding his book about infinities... It really is great that some of Bolzano's texts are available to our day :D They are quite inspirational.
– mathreadler
Jul 28 at 18:16
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
It is not surprising that there is no constructive way to identify, at each stage, which of the $2^n$ boxes holds infinitely many points—because, from the outset, the premiss of the theorem assumes nothing about where the points are: we know only that there are infinitely many of them. There are thus at least continuum-many possibilities for how the points might be distributed. All we can do is make the logical observation that, at every stage, the boxes cannot all hold a finite number of points, and therefore at least one of them holds an infinite number. If we need to identify just which such a box might be, we might be able to do so if we were given a precise construction of $A$; but we are given no such information.
answered Jul 28 at 8:45
John Bentin
10.8k22350
add a comment |Â
up vote
7
down vote
accepted
It is not surprising that there is no constructive way to identify, at each stage, which of the $2^n$ boxes holds infinitely many points—because, from the outset, the premiss of the theorem assumes nothing about where the points are: we know only that there are infinitely many of them. There are thus at least continuum-many possibilities for how the points might be distributed. All we can do is make the logical observation that, at every stage, the boxes cannot all hold a finite number of points, and therefore at least one of them holds an infinite number. If we need to identify just which such a box might be, we might be able to do so if we were given a precise construction of $A$; but we are given no such information.
answered Jul 28 at 8:45
John Bentin
10.8k22350
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
It is not surprising that there is no constructive way to identify, at each stage, which of the $2^n$ boxes holds infinitely many points—because, from the outset, the premiss of the theorem assumes nothing about where the points are: we know only that there are infinitely many of them. There are thus at least continuum-many possibilities for how the points might be distributed. All we can do is make the logical observation that, at every stage, the boxes cannot all hold a finite number of points, and therefore at least one of them holds an infinite number. If we need to identify just which such a box might be, we might be able to do so if we were given a precise construction of $A$; but we are given no such information.
answered Jul 28 at 8:45
John Bentin
10.8k22350
It is not surprising that there is no constructive way to identify, at each stage, which of the $2^n$ boxes holds infinitely many points—because, from the outset, the premiss of the theorem assumes nothing about where the points are: we know only that there are infinitely many of them. There are thus at least continuum-many possibilities for how the points might be distributed. All we can do is make the logical observation that, at every stage, the boxes cannot all hold a finite number of points, and therefore at least one of them holds an infinite number. If we need to identify just which such a box might be, we might be able to do so if we were given a precise construction of $A$; but we are given no such information.
answered Jul 28 at 8:45
John Bentin
10.8k22350
answered Jul 28 at 8:45
John Bentin
10.8k22350
answered Jul 28 at 8:45
John Bentin
10.8k22350
answered Jul 28 at 8:45
John Bentin
10.8k22350
10.8k22350
add a comment |Â
add a comment |Â
up vote
7
down vote
It seems that you understand why at least one box contains infinitely many elements of $A$, and your difficulty is only in finding which box that is (or finding one such box if there are several of them). That is, it's a question of existence versus finding. Notice, though, that the theorem you're proving says only that there exists a limit point. It doesn't claim anything about finding a limit point.
To prove the existence of a limit point, it's enough to have the existence of an appropriate box at every stage of the process.
If you wanted to prove a stronger theorem saying that one can find a limit point, then you'd have to worry about finding an appropriate box at every stage. Any alleged theorem of that sort would have to be preceded by information about (1) exactly how the set $A$ is presented and (2) exactly what it means to "find" a point. So such a theorem would not only be harder to prove but considerably more complicated to state than the Bolzano-Weierstrass theorem.
answered Jul 28 at 12:51
Andreas Blass
47.5k348104
add a comment |Â
up vote
7
down vote
It seems that you understand why at least one box contains infinitely many elements of $A$, and your difficulty is only in finding which box that is (or finding one such box if there are several of them). That is, it's a question of existence versus finding. Notice, though, that the theorem you're proving says only that there exists a limit point. It doesn't claim anything about finding a limit point.
To prove the existence of a limit point, it's enough to have the existence of an appropriate box at every stage of the process.
If you wanted to prove a stronger theorem saying that one can find a limit point, then you'd have to worry about finding an appropriate box at every stage. Any alleged theorem of that sort would have to be preceded by information about (1) exactly how the set $A$ is presented and (2) exactly what it means to "find" a point. So such a theorem would not only be harder to prove but considerably more complicated to state than the Bolzano-Weierstrass theorem.
answered Jul 28 at 12:51
Andreas Blass
47.5k348104
add a comment |Â
up vote
7
down vote
up vote
7
down vote
It seems that you understand why at least one box contains infinitely many elements of $A$, and your difficulty is only in finding which box that is (or finding one such box if there are several of them). That is, it's a question of existence versus finding. Notice, though, that the theorem you're proving says only that there exists a limit point. It doesn't claim anything about finding a limit point.
To prove the existence of a limit point, it's enough to have the existence of an appropriate box at every stage of the process.
If you wanted to prove a stronger theorem saying that one can find a limit point, then you'd have to worry about finding an appropriate box at every stage. Any alleged theorem of that sort would have to be preceded by information about (1) exactly how the set $A$ is presented and (2) exactly what it means to "find" a point. So such a theorem would not only be harder to prove but considerably more complicated to state than the Bolzano-Weierstrass theorem.
answered Jul 28 at 12:51
Andreas Blass
47.5k348104
It seems that you understand why at least one box contains infinitely many elements of $A$, and your difficulty is only in finding which box that is (or finding one such box if there are several of them). That is, it's a question of existence versus finding. Notice, though, that the theorem you're proving says only that there exists a limit point. It doesn't claim anything about finding a limit point.
To prove the existence of a limit point, it's enough to have the existence of an appropriate box at every stage of the process.
If you wanted to prove a stronger theorem saying that one can find a limit point, then you'd have to worry about finding an appropriate box at every stage. Any alleged theorem of that sort would have to be preceded by information about (1) exactly how the set $A$ is presented and (2) exactly what it means to "find" a point. So such a theorem would not only be harder to prove but considerably more complicated to state than the Bolzano-Weierstrass theorem.
answered Jul 28 at 12:51
Andreas Blass
47.5k348104
answered Jul 28 at 12:51
Andreas Blass
47.5k348104
answered Jul 28 at 12:51
Andreas Blass
47.5k348104
answered Jul 28 at 12:51
Andreas Blass
47.5k348104
47.5k348104
add a comment |Â
add a comment |Â
up vote
4
down vote
It is in fact a philosophical question. The "paradoxes of infinity" have puzzled people since ancient times. In 1851 Bolzano himself has written a book with the title "Paradoxien des Unendlichen" (in English "Paradoxes of the Infinite").
In your proof you start with an infinite set $A$. But how can you know that is infinite? Is it explicitly given by listing its elements? In broad terms let us understand by "listing" any rule that allows you to decide whether a given element $x$ belongs to $A$ or not.
If you have such a listing, then you will be able to identify $B_2, B_3, ...$. If not, you are not even able to identify $A$. In that case it is only possible to argue on an abstract level: If you put infinitely many things into finitely many boxes, than at least one box must contain infinitely many things. Imagine a "person" has put the things into the boxes, but you didn't watch the process. Then you will not know which box contains infinitely many things, but you will know that at least one box must contain infinitely many things. If you don't like to consider "infinitely many things", you can also consider $10$ things in $2$ boxes: You know that at least one must contain $5$ things or more, but you don't know which box. It is similar to the shell game: You know that the ball is placed beneath one the shells, but would you bet beneath which?
answered Jul 28 at 8:55
Paul Frost
3,603420
Regarding his book about infinities... It really is great that some of Bolzano's texts are available to our day :D They are quite inspirational.
– mathreadler
Jul 28 at 18:16
add a comment |Â
up vote
4
down vote
It is in fact a philosophical question. The "paradoxes of infinity" have puzzled people since ancient times. In 1851 Bolzano himself has written a book with the title "Paradoxien des Unendlichen" (in English "Paradoxes of the Infinite").
In your proof you start with an infinite set $A$. But how can you know that is infinite? Is it explicitly given by listing its elements? In broad terms let us understand by "listing" any rule that allows you to decide whether a given element $x$ belongs to $A$ or not.
If you have such a listing, then you will be able to identify $B_2, B_3, ...$. If not, you are not even able to identify $A$. In that case it is only possible to argue on an abstract level: If you put infinitely many things into finitely many boxes, than at least one box must contain infinitely many things. Imagine a "person" has put the things into the boxes, but you didn't watch the process. Then you will not know which box contains infinitely many things, but you will know that at least one box must contain infinitely many things. If you don't like to consider "infinitely many things", you can also consider $10$ things in $2$ boxes: You know that at least one must contain $5$ things or more, but you don't know which box. It is similar to the shell game: You know that the ball is placed beneath one the shells, but would you bet beneath which?
answered Jul 28 at 8:55
Paul Frost
3,603420
Regarding his book about infinities... It really is great that some of Bolzano's texts are available to our day :D They are quite inspirational.
– mathreadler
Jul 28 at 18:16
add a comment |Â
up vote
4
down vote
up vote
4
down vote
It is in fact a philosophical question. The "paradoxes of infinity" have puzzled people since ancient times. In 1851 Bolzano himself has written a book with the title "Paradoxien des Unendlichen" (in English "Paradoxes of the Infinite").
In your proof you start with an infinite set $A$. But how can you know that is infinite? Is it explicitly given by listing its elements? In broad terms let us understand by "listing" any rule that allows you to decide whether a given element $x$ belongs to $A$ or not.
If you have such a listing, then you will be able to identify $B_2, B_3, ...$. If not, you are not even able to identify $A$. In that case it is only possible to argue on an abstract level: If you put infinitely many things into finitely many boxes, than at least one box must contain infinitely many things. Imagine a "person" has put the things into the boxes, but you didn't watch the process. Then you will not know which box contains infinitely many things, but you will know that at least one box must contain infinitely many things. If you don't like to consider "infinitely many things", you can also consider $10$ things in $2$ boxes: You know that at least one must contain $5$ things or more, but you don't know which box. It is similar to the shell game: You know that the ball is placed beneath one the shells, but would you bet beneath which?
answered Jul 28 at 8:55
Paul Frost
3,603420
It is in fact a philosophical question. The "paradoxes of infinity" have puzzled people since ancient times. In 1851 Bolzano himself has written a book with the title "Paradoxien des Unendlichen" (in English "Paradoxes of the Infinite").
In your proof you start with an infinite set $A$. But how can you know that is infinite? Is it explicitly given by listing its elements? In broad terms let us understand by "listing" any rule that allows you to decide whether a given element $x$ belongs to $A$ or not.
If you have such a listing, then you will be able to identify $B_2, B_3, ...$. If not, you are not even able to identify $A$. In that case it is only possible to argue on an abstract level: If you put infinitely many things into finitely many boxes, than at least one box must contain infinitely many things. Imagine a "person" has put the things into the boxes, but you didn't watch the process. Then you will not know which box contains infinitely many things, but you will know that at least one box must contain infinitely many things. If you don't like to consider "infinitely many things", you can also consider $10$ things in $2$ boxes: You know that at least one must contain $5$ things or more, but you don't know which box. It is similar to the shell game: You know that the ball is placed beneath one the shells, but would you bet beneath which?
answered Jul 28 at 8:55
Paul Frost
3,603420
edited Jul 28 at 10:56
answered Jul 28 at 8:55
Paul Frost
3,603420
answered Jul 28 at 8:55
Paul Frost
3,603420
answered Jul 28 at 8:55
Paul Frost
3,603420
3,603420
Regarding his book about infinities... It really is great that some of Bolzano's texts are available to our day :D They are quite inspirational.
– mathreadler
Jul 28 at 18:16
add a comment |Â
Regarding his book about infinities... It really is great that some of Bolzano's texts are available to our day :D They are quite inspirational.
– mathreadler
Jul 28 at 18:16
Regarding his book about infinities... It really is great that some of Bolzano's texts are available to our day :D They are quite inspirational.
– mathreadler
Jul 28 at 18:16
Regarding his book about infinities... It really is great that some of Bolzano's texts are available to our day :D They are quite inspirational.
– mathreadler
Jul 28 at 18:16
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You say at least one sub-box has. Also there is a missing step somewhere, the highest $B_k$ mentioned is $B_2$ but you then write $xin cap_nB_n$ we don't know what any $B_k, k>2$ is at this point.
– mathreadler
Jul 28 at 7:53
3
What do you mean exactly by "identify"? And why do we need to do so?
– Anguepa
Jul 28 at 7:53
@Angeupa What I mean by "identify" is, what we need to do before "Let's say that sub-box as $B_2$". I can prove at least one such set exists, but how can we actually pick that set out and label it as $B_2$?
– moreblue
Jul 28 at 8:01
4
It is an existence proof not a construction proof. It only shows that such a point must exist, not that there must be only one or how to find it (if it is only one). The points must sum up and the only way to sum to infinity is if one or more of the subboxes sums to infinity since if none of them did then it would be finite number of points which would violate the assumption.
– mathreadler
Jul 28 at 8:04
Since there are finitely many sub-boxes ($2^N$), one can (in principle) check each box until we find one that contains infinitely many elements.
– Brahadeesh
Jul 28 at 8:04