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Converse of Bolzano-Weierstrass theorem
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Bolzano-Weierstrass theorem states that every bounded sequence has a limit point. But, the converse is not true.
That is, there are some unbounded sequences which have a limit point. In my course book, I found an example for this claim, but it doesn't make sense.
Here's the example give in the book:
The set: 1, 2, 1, 4, 1, 6, ... is unbounded, but has a limit point of 1.
I can't understand how this set has a limit point as 1. According to the book definition of limit point, 'x' is the limit point of a sequence, if every neighborhood of 'x' has infinitely many elements of the sequence. If I apply it here, then I get only infinity as the limit point. Am I missing something?
real-analysis sequences-and-series limits
edited 36 mins ago
Eric Wofsey
169k12197314
asked 49 mins ago
Sankalp
1084
New contributor
Sankalp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
1
down vote
favorite
Bolzano-Weierstrass theorem states that every bounded sequence has a limit point. But, the converse is not true.
That is, there are some unbounded sequences which have a limit point. In my course book, I found an example for this claim, but it doesn't make sense.
Here's the example give in the book:
The set: 1, 2, 1, 4, 1, 6, ... is unbounded, but has a limit point of 1.
I can't understand how this set has a limit point as 1. According to the book definition of limit point, 'x' is the limit point of a sequence, if every neighborhood of 'x' has infinitely many elements of the sequence. If I apply it here, then I get only infinity as the limit point. Am I missing something?
real-analysis sequences-and-series limits
edited 36 mins ago
Eric Wofsey
169k12197314
asked 49 mins ago
Sankalp
1084
New contributor
Sankalp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
11 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Bolzano-Weierstrass theorem states that every bounded sequence has a limit point. But, the converse is not true.
That is, there are some unbounded sequences which have a limit point. In my course book, I found an example for this claim, but it doesn't make sense.
Here's the example give in the book:
The set: 1, 2, 1, 4, 1, 6, ... is unbounded, but has a limit point of 1.
I can't understand how this set has a limit point as 1. According to the book definition of limit point, 'x' is the limit point of a sequence, if every neighborhood of 'x' has infinitely many elements of the sequence. If I apply it here, then I get only infinity as the limit point. Am I missing something?
real-analysis sequences-and-series limits
edited 36 mins ago
Eric Wofsey
169k12197314
asked 49 mins ago
Sankalp
1084
New contributor
Sankalp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Bolzano-Weierstrass theorem states that every bounded sequence has a limit point. But, the converse is not true.
That is, there are some unbounded sequences which have a limit point. In my course book, I found an example for this claim, but it doesn't make sense.
Here's the example give in the book:
The set: 1, 2, 1, 4, 1, 6, ... is unbounded, but has a limit point of 1.
I can't understand how this set has a limit point as 1. According to the book definition of limit point, 'x' is the limit point of a sequence, if every neighborhood of 'x' has infinitely many elements of the sequence. If I apply it here, then I get only infinity as the limit point. Am I missing something?
real-analysis sequences-and-series limits
real-analysis sequences-and-series limits
edited 36 mins ago
Eric Wofsey
169k12197314
asked 49 mins ago
Sankalp
1084
New contributor
Sankalp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 36 mins ago
Eric Wofsey
169k12197314
asked 49 mins ago
Sankalp
1084
New contributor
Sankalp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 36 mins ago
Eric Wofsey
169k12197314
edited 36 mins ago
Eric Wofsey
169k12197314
edited 36 mins ago
Eric Wofsey
169k12197314
169k12197314
asked 49 mins ago
Sankalp
1084
New contributor
Sankalp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 49 mins ago
Sankalp
1084
asked 49 mins ago
Sankalp
1084
1084
New contributor
Sankalp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sankalp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sankalp is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
11 mins ago
add a comment |Â
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
11 mins ago
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
11 mins ago
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
11 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
It seems there is some confusion between sets and sequences. In this example, what you have written as "1, 2, 1, 4, 1, 6, ..." is not meant to be a set but rather a sequence $(a_n)$ with $a_0=1,a_1=2,a_2=1,a_3=4,dots$.
In particular, then, when we say a point $x$ is a limit point of $(a_n)$, this means that for every neighborhood $U$ of $x$, there exist infinitely many $ninmathbbN$ such that $a_nin U$. It does not mean there are infinitely many different numbers in the sequence which are in $U$, since these values of $a_n$ for different $n$ might actually be the same. So in this case, since every neighborhood of $1$ contains $1$, it contains $a_0,a_2,a_4,dots$, and so $1$ is a limit point of the sequence.
(In contrast, $1$ is not a limit point of the set $1, 2, 1, 4, 1, 6, dots=1, 2, 4, 6, dots$ because $(0,2)$ is a neighborhood of $1$ that contains only one element of this set, namely $1$.)
answered 37 mins ago
Eric Wofsey
169k12197314
Thanks for clearing my confusion! It was indeed between sets and sequences.
– Sankalp
33 mins ago
Eric, very clear and nice as usual
– Peter Szilas
30 mins ago
add a comment |Â
up vote
2
down vote
We have
$$a_n = begincases 1, & n text is odd \ n, & n text is evenendcases$$
$1$ appears infinitely often. Hence there is a subsequence that always take value $1$. That subsequence converges to $1$. Hence $1$ is a limit to the subsequence.
answered 47 mins ago
Siong Thye Goh
85k1457107
But, according to the limit point definition, 'x' is the limit point of a sequence, if every neighborhood of 'x' has infinitely many elements of the sequence.
– Sankalp
44 mins ago
1
I think Bolzano states that every bounded sequence has a convergence subsequence.
– Siong Thye Goh
40 mins ago
The third paragraph of wikipedia might be of interest to you
– Siong Thye Goh
37 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It seems there is some confusion between sets and sequences. In this example, what you have written as "1, 2, 1, 4, 1, 6, ..." is not meant to be a set but rather a sequence $(a_n)$ with $a_0=1,a_1=2,a_2=1,a_3=4,dots$.
In particular, then, when we say a point $x$ is a limit point of $(a_n)$, this means that for every neighborhood $U$ of $x$, there exist infinitely many $ninmathbbN$ such that $a_nin U$. It does not mean there are infinitely many different numbers in the sequence which are in $U$, since these values of $a_n$ for different $n$ might actually be the same. So in this case, since every neighborhood of $1$ contains $1$, it contains $a_0,a_2,a_4,dots$, and so $1$ is a limit point of the sequence.
(In contrast, $1$ is not a limit point of the set $1, 2, 1, 4, 1, 6, dots=1, 2, 4, 6, dots$ because $(0,2)$ is a neighborhood of $1$ that contains only one element of this set, namely $1$.)
answered 37 mins ago
Eric Wofsey
169k12197314
Thanks for clearing my confusion! It was indeed between sets and sequences.
– Sankalp
33 mins ago
Eric, very clear and nice as usual
– Peter Szilas
30 mins ago
add a comment |Â
up vote
2
down vote
accepted
It seems there is some confusion between sets and sequences. In this example, what you have written as "1, 2, 1, 4, 1, 6, ..." is not meant to be a set but rather a sequence $(a_n)$ with $a_0=1,a_1=2,a_2=1,a_3=4,dots$.
In particular, then, when we say a point $x$ is a limit point of $(a_n)$, this means that for every neighborhood $U$ of $x$, there exist infinitely many $ninmathbbN$ such that $a_nin U$. It does not mean there are infinitely many different numbers in the sequence which are in $U$, since these values of $a_n$ for different $n$ might actually be the same. So in this case, since every neighborhood of $1$ contains $1$, it contains $a_0,a_2,a_4,dots$, and so $1$ is a limit point of the sequence.
(In contrast, $1$ is not a limit point of the set $1, 2, 1, 4, 1, 6, dots=1, 2, 4, 6, dots$ because $(0,2)$ is a neighborhood of $1$ that contains only one element of this set, namely $1$.)
answered 37 mins ago
Eric Wofsey
169k12197314
Thanks for clearing my confusion! It was indeed between sets and sequences.
– Sankalp
33 mins ago
Eric, very clear and nice as usual
– Peter Szilas
30 mins ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It seems there is some confusion between sets and sequences. In this example, what you have written as "1, 2, 1, 4, 1, 6, ..." is not meant to be a set but rather a sequence $(a_n)$ with $a_0=1,a_1=2,a_2=1,a_3=4,dots$.
In particular, then, when we say a point $x$ is a limit point of $(a_n)$, this means that for every neighborhood $U$ of $x$, there exist infinitely many $ninmathbbN$ such that $a_nin U$. It does not mean there are infinitely many different numbers in the sequence which are in $U$, since these values of $a_n$ for different $n$ might actually be the same. So in this case, since every neighborhood of $1$ contains $1$, it contains $a_0,a_2,a_4,dots$, and so $1$ is a limit point of the sequence.
(In contrast, $1$ is not a limit point of the set $1, 2, 1, 4, 1, 6, dots=1, 2, 4, 6, dots$ because $(0,2)$ is a neighborhood of $1$ that contains only one element of this set, namely $1$.)
answered 37 mins ago
Eric Wofsey
169k12197314
It seems there is some confusion between sets and sequences. In this example, what you have written as "1, 2, 1, 4, 1, 6, ..." is not meant to be a set but rather a sequence $(a_n)$ with $a_0=1,a_1=2,a_2=1,a_3=4,dots$.
In particular, then, when we say a point $x$ is a limit point of $(a_n)$, this means that for every neighborhood $U$ of $x$, there exist infinitely many $ninmathbbN$ such that $a_nin U$. It does not mean there are infinitely many different numbers in the sequence which are in $U$, since these values of $a_n$ for different $n$ might actually be the same. So in this case, since every neighborhood of $1$ contains $1$, it contains $a_0,a_2,a_4,dots$, and so $1$ is a limit point of the sequence.
(In contrast, $1$ is not a limit point of the set $1, 2, 1, 4, 1, 6, dots=1, 2, 4, 6, dots$ because $(0,2)$ is a neighborhood of $1$ that contains only one element of this set, namely $1$.)
answered 37 mins ago
Eric Wofsey
169k12197314
answered 37 mins ago
Eric Wofsey
169k12197314
answered 37 mins ago
Eric Wofsey
169k12197314
answered 37 mins ago
Eric Wofsey
169k12197314
169k12197314
Thanks for clearing my confusion! It was indeed between sets and sequences.
– Sankalp
33 mins ago
Eric, very clear and nice as usual
– Peter Szilas
30 mins ago
add a comment |Â
Thanks for clearing my confusion! It was indeed between sets and sequences.
– Sankalp
33 mins ago
Eric, very clear and nice as usual
– Peter Szilas
30 mins ago
Thanks for clearing my confusion! It was indeed between sets and sequences.
– Sankalp
33 mins ago
Thanks for clearing my confusion! It was indeed between sets and sequences.
– Sankalp
33 mins ago
Eric, very clear and nice as usual
– Peter Szilas
30 mins ago
Eric, very clear and nice as usual
– Peter Szilas
30 mins ago
add a comment |Â
up vote
2
down vote
We have
$$a_n = begincases 1, & n text is odd \ n, & n text is evenendcases$$
$1$ appears infinitely often. Hence there is a subsequence that always take value $1$. That subsequence converges to $1$. Hence $1$ is a limit to the subsequence.
answered 47 mins ago
Siong Thye Goh
85k1457107
But, according to the limit point definition, 'x' is the limit point of a sequence, if every neighborhood of 'x' has infinitely many elements of the sequence.
– Sankalp
44 mins ago
1
I think Bolzano states that every bounded sequence has a convergence subsequence.
– Siong Thye Goh
40 mins ago
The third paragraph of wikipedia might be of interest to you
– Siong Thye Goh
37 mins ago
add a comment |Â
up vote
2
down vote
We have
$$a_n = begincases 1, & n text is odd \ n, & n text is evenendcases$$
$1$ appears infinitely often. Hence there is a subsequence that always take value $1$. That subsequence converges to $1$. Hence $1$ is a limit to the subsequence.
answered 47 mins ago
Siong Thye Goh
85k1457107
But, according to the limit point definition, 'x' is the limit point of a sequence, if every neighborhood of 'x' has infinitely many elements of the sequence.
– Sankalp
44 mins ago
1
I think Bolzano states that every bounded sequence has a convergence subsequence.
– Siong Thye Goh
40 mins ago
The third paragraph of wikipedia might be of interest to you
– Siong Thye Goh
37 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We have
$$a_n = begincases 1, & n text is odd \ n, & n text is evenendcases$$
$1$ appears infinitely often. Hence there is a subsequence that always take value $1$. That subsequence converges to $1$. Hence $1$ is a limit to the subsequence.
answered 47 mins ago
Siong Thye Goh
85k1457107
We have
$$a_n = begincases 1, & n text is odd \ n, & n text is evenendcases$$
$1$ appears infinitely often. Hence there is a subsequence that always take value $1$. That subsequence converges to $1$. Hence $1$ is a limit to the subsequence.
answered 47 mins ago
Siong Thye Goh
85k1457107
edited 41 mins ago
answered 47 mins ago
Siong Thye Goh
85k1457107
answered 47 mins ago
Siong Thye Goh
85k1457107
answered 47 mins ago
Siong Thye Goh
85k1457107
85k1457107
But, according to the limit point definition, 'x' is the limit point of a sequence, if every neighborhood of 'x' has infinitely many elements of the sequence.
– Sankalp
44 mins ago
1
I think Bolzano states that every bounded sequence has a convergence subsequence.
– Siong Thye Goh
40 mins ago
The third paragraph of wikipedia might be of interest to you
– Siong Thye Goh
37 mins ago
add a comment |Â
But, according to the limit point definition, 'x' is the limit point of a sequence, if every neighborhood of 'x' has infinitely many elements of the sequence.
– Sankalp
44 mins ago
1
I think Bolzano states that every bounded sequence has a convergence subsequence.
– Siong Thye Goh
40 mins ago
The third paragraph of wikipedia might be of interest to you
– Siong Thye Goh
37 mins ago
But, according to the limit point definition, 'x' is the limit point of a sequence, if every neighborhood of 'x' has infinitely many elements of the sequence.
– Sankalp
44 mins ago
But, according to the limit point definition, 'x' is the limit point of a sequence, if every neighborhood of 'x' has infinitely many elements of the sequence.
– Sankalp
44 mins ago
1
1
I think Bolzano states that every bounded sequence has a convergence subsequence.
– Siong Thye Goh
40 mins ago
I think Bolzano states that every bounded sequence has a convergence subsequence.
– Siong Thye Goh
40 mins ago
The third paragraph of wikipedia might be of interest to you
– Siong Thye Goh
37 mins ago
The third paragraph of wikipedia might be of interest to you
– Siong Thye Goh
37 mins ago
add a comment |Â
Sankalp is a new contributor. Be nice, and check out our Code of Conduct.
Sankalp is a new contributor. Be nice, and check out our Code of Conduct.
Sankalp is a new contributor. Be nice, and check out our Code of Conduct.
Sankalp is a new contributor. Be nice, and check out our Code of Conduct.
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
11 mins ago