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Sequence with convergent subnets but no convergent subsequences

Product space that is compact, but isn't sequentially compactSubnets vs. SubsequencesExample of converging subnet, when there is no converging subsequenceA bounded net with a unique limit point must be convergentSubnet vs SubsequenceIs a sequentially compact (non-metrizable) uniform space totally bounded?Proper maps and convergent subnetsHow to build a subnet out of these subnets?Trying to understand how a subnet of a sequence differs from a subsequenceSequential compactness vs compactnessHow is “every net in $X$ has a convergent subnet” different from sequential compactness?

1

1

$begingroup$

We can regard a sequence as a special kind of net. But the definition of "subnet" is more flexible than that of "subsequence", so it's easy to find subnets of a sequence that aren't subsequences.

In fact, if $X$ is a compact topological space that is not sequentially compact, like

$$ X = prod_x in mathbbR [0,1] ; ,$$

we can have a sequence in $X$ that has no convergent subsequences, but it must have convergent subnets! I've always found this phenomenon mysterious.

Can someone describe, as explicitly as possible, a sequence in some topological space that has no convergent subsequences, but has a convergent subnet?

Does finding an example require the axiom of choice, or is there an 'explicit' one?

general-topology analysis

share|cite|improve this question

edited Mar 22 at 1:22

John Baez

asked Mar 21 at 18:22

John BaezJohn Baez

33619

$endgroup$

add a comment | 

1

1

$begingroup$

We can regard a sequence as a special kind of net. But the definition of "subnet" is more flexible than that of "subsequence", so it's easy to find subnets of a sequence that aren't subsequences.

In fact, if $X$ is a compact topological space that is not sequentially compact, like

$$ X = prod_x in mathbbR [0,1] ; ,$$

we can have a sequence in $X$ that has no convergent subsequences, but it must have convergent subnets! I've always found this phenomenon mysterious.

Can someone describe, as explicitly as possible, a sequence in some topological space that has no convergent subsequences, but has a convergent subnet?

Does finding an example require the axiom of choice, or is there an 'explicit' one?

general-topology analysis

share|cite|improve this question

edited Mar 22 at 1:22

John Baez

asked Mar 21 at 18:22

John BaezJohn Baez

33619

$endgroup$

add a comment | 

$begingroup$

We can regard a sequence as a special kind of net. But the definition of "subnet" is more flexible than that of "subsequence", so it's easy to find subnets of a sequence that aren't subsequences.

In fact, if $X$ is a compact topological space that is not sequentially compact, like

$$ X = prod_x in mathbbR [0,1] ; ,$$

we can have a sequence in $X$ that has no convergent subsequences, but it must have convergent subnets! I've always found this phenomenon mysterious.

Can someone describe, as explicitly as possible, a sequence in some topological space that has no convergent subsequences, but has a convergent subnet?

Does finding an example require the axiom of choice, or is there an 'explicit' one?

general-topology analysis

share|cite|improve this question

edited Mar 22 at 1:22

John Baez

asked Mar 21 at 18:22

John BaezJohn Baez

33619

$endgroup$

share|cite|improve this question

edited Mar 22 at 1:22

John Baez

asked Mar 21 at 18:22

John BaezJohn Baez

33619

share|cite|improve this question

edited Mar 22 at 1:22

John Baez

asked Mar 21 at 18:22

John BaezJohn Baez

33619

asked Mar 21 at 18:22

John BaezJohn Baez

33619

asked Mar 21 at 18:22

John BaezJohn Baez

33619

1 Answer
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oldest

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3

$begingroup$

I give an explicit example here: let $X = 0,1^I$ where $I = 0,1^mathbbN$. This is a compact space by Tychonoff's theorem, so every net has a convergent subnet.

If we denote for $i in I$ and $n in mathbbN$ by $pi_n(i)$ the $n$-th coordinate of the sequence (or function) $i$, then the required sequence is $(f_n)_n$, where all $f_n : I to 0,1$ are given by $f_n(i) = pi_n(i)$ for all $i in I$.

In the linked answer I give a diagonalisation argument why no subsequence of $(f_n)$ can converge in $X$ (i.e. pointwise).

I think a convergent subnet of the $(f_n)$ (which exists by compactness) will probably involve some ultrafilter on $mathbbN$, e.g. and so won't be as explicit.

share|cite|improve this answer

answered Mar 21 at 22:21

Henno BrandsmaHenno Brandsma

115k349125

$endgroup$

add a comment | 

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3

$begingroup$

I give an explicit example here: let $X = 0,1^I$ where $I = 0,1^mathbbN$. This is a compact space by Tychonoff's theorem, so every net has a convergent subnet.

If we denote for $i in I$ and $n in mathbbN$ by $pi_n(i)$ the $n$-th coordinate of the sequence (or function) $i$, then the required sequence is $(f_n)_n$, where all $f_n : I to 0,1$ are given by $f_n(i) = pi_n(i)$ for all $i in I$.

In the linked answer I give a diagonalisation argument why no subsequence of $(f_n)$ can converge in $X$ (i.e. pointwise).

I think a convergent subnet of the $(f_n)$ (which exists by compactness) will probably involve some ultrafilter on $mathbbN$, e.g. and so won't be as explicit.

share|cite|improve this answer

answered Mar 21 at 22:21

Henno BrandsmaHenno Brandsma

115k349125

$endgroup$

add a comment | 

3

$begingroup$

I give an explicit example here: let $X = 0,1^I$ where $I = 0,1^mathbbN$. This is a compact space by Tychonoff's theorem, so every net has a convergent subnet.

If we denote for $i in I$ and $n in mathbbN$ by $pi_n(i)$ the $n$-th coordinate of the sequence (or function) $i$, then the required sequence is $(f_n)_n$, where all $f_n : I to 0,1$ are given by $f_n(i) = pi_n(i)$ for all $i in I$.

In the linked answer I give a diagonalisation argument why no subsequence of $(f_n)$ can converge in $X$ (i.e. pointwise).

I think a convergent subnet of the $(f_n)$ (which exists by compactness) will probably involve some ultrafilter on $mathbbN$, e.g. and so won't be as explicit.

share|cite|improve this answer

answered Mar 21 at 22:21

Henno BrandsmaHenno Brandsma

115k349125

$endgroup$

add a comment | 

$begingroup$

I give an explicit example here: let $X = 0,1^I$ where $I = 0,1^mathbbN$. This is a compact space by Tychonoff's theorem, so every net has a convergent subnet.

If we denote for $i in I$ and $n in mathbbN$ by $pi_n(i)$ the $n$-th coordinate of the sequence (or function) $i$, then the required sequence is $(f_n)_n$, where all $f_n : I to 0,1$ are given by $f_n(i) = pi_n(i)$ for all $i in I$.

In the linked answer I give a diagonalisation argument why no subsequence of $(f_n)$ can converge in $X$ (i.e. pointwise).

I think a convergent subnet of the $(f_n)$ (which exists by compactness) will probably involve some ultrafilter on $mathbbN$, e.g. and so won't be as explicit.

share|cite|improve this answer

answered Mar 21 at 22:21

Henno BrandsmaHenno Brandsma

115k349125

$endgroup$

share|cite|improve this answer

answered Mar 21 at 22:21

Henno BrandsmaHenno Brandsma

115k349125

answered Mar 21 at 22:21

Henno BrandsmaHenno Brandsma

115k349125

answered Mar 21 at 22:21

Henno BrandsmaHenno Brandsma

115k349125