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is it always possible to make a converging sequence to a limit point?

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0

1

$begingroup$

X is a topology; p $notin$ A $subset$ X;

if p is a limit point of A; that is, every [open subset of X containing p] has non-empty intersection with A,

then does there always exist a sequence of A that converges to p?

this is true if X is a metric space, however i am not sure whether this is true or not in general cases.

if it's false, please give me a counterexample.

if this is true, does the proof require any sort of choice-related axioms other than ZFC's?

one more thing;

f : X $rightarrow$ Y, where X and Y are topologies; x $in$ X;

assume that for every [sequence x$_n$ of X that converges to x], f(x$_n$) converges to f(x).

then is f continuous on x?; that is, for every [open subset U $subset$ Y, containing f(x)] there exists an open subset V $subset$ X s.t. f(V) $subset$ U

if X is a metric space, one can prove the above by contradiction, making a sequence that converges to x but its image not converging to f(x).

is this true in general?

general-topology set-theory

share|cite|improve this question

edited Mar 21 at 5:16

JHL

asked Mar 21 at 5:03

JHLJHL

34

$endgroup$

add a comment | 

0

1

$begingroup$

X is a topology; p $notin$ A $subset$ X;

if p is a limit point of A; that is, every [open subset of X containing p] has non-empty intersection with A,

then does there always exist a sequence of A that converges to p?

this is true if X is a metric space, however i am not sure whether this is true or not in general cases.

if it's false, please give me a counterexample.

if this is true, does the proof require any sort of choice-related axioms other than ZFC's?

one more thing;

f : X $rightarrow$ Y, where X and Y are topologies; x $in$ X;

assume that for every [sequence x$_n$ of X that converges to x], f(x$_n$) converges to f(x).

then is f continuous on x?; that is, for every [open subset U $subset$ Y, containing f(x)] there exists an open subset V $subset$ X s.t. f(V) $subset$ U

if X is a metric space, one can prove the above by contradiction, making a sequence that converges to x but its image not converging to f(x).

is this true in general?

general-topology set-theory

share|cite|improve this question

edited Mar 21 at 5:16

JHL

asked Mar 21 at 5:03

JHLJHL

34

$endgroup$

add a comment | 

$begingroup$

X is a topology; p $notin$ A $subset$ X;

if p is a limit point of A; that is, every [open subset of X containing p] has non-empty intersection with A,

then does there always exist a sequence of A that converges to p?

this is true if X is a metric space, however i am not sure whether this is true or not in general cases.

if it's false, please give me a counterexample.

if this is true, does the proof require any sort of choice-related axioms other than ZFC's?

one more thing;

f : X $rightarrow$ Y, where X and Y are topologies; x $in$ X;

assume that for every [sequence x$_n$ of X that converges to x], f(x$_n$) converges to f(x).

then is f continuous on x?; that is, for every [open subset U $subset$ Y, containing f(x)] there exists an open subset V $subset$ X s.t. f(V) $subset$ U

if X is a metric space, one can prove the above by contradiction, making a sequence that converges to x but its image not converging to f(x).

is this true in general?

general-topology set-theory

share|cite|improve this question

edited Mar 21 at 5:16

JHL

asked Mar 21 at 5:03

JHLJHL

34

$endgroup$

share|cite|improve this question

edited Mar 21 at 5:16

JHL

asked Mar 21 at 5:03

JHLJHL

34

share|cite|improve this question

edited Mar 21 at 5:16

JHL

asked Mar 21 at 5:03

JHLJHL

34

asked Mar 21 at 5:03

JHLJHL

34

asked Mar 21 at 5:03

JHLJHL

34

1 Answer
1

active

oldest

votes

2

$begingroup$

It's true in a metric space, as a metric space is first countable and we can use countable choice to show it (well within ZFC).

It's not true in general: let $X$ be the reals (or any uncountable set) in the co-countable topology. Any sequence in $X$ that converges to $p$ is equal to $p$ after finitely many terms (eventually constant); many proofs exist for this on this site.

So $0 in overlineA$ where $A= (0,infty)$ but no sequence in $A$ converges to $0$. Also, the identity map from $X$ to $Y$, the reals in the usual topology, is not continuous but trivially sequentially continuous.

share|cite|improve this answer

answered Mar 21 at 5:24

Henno BrandsmaHenno Brandsma

115k349125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I may be mistaken as I'm certainly not a topologist, but can't these statements be generalized in terms of nets?
  $endgroup$
  – Gary Moon
  Mar 21 at 5:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GaryMoon sure, there is always a net converging from $A$ to $p$ and continuity is net-convergence preservation. But the OP asked for sequences.
  $endgroup$
  – Henno Brandsma
  Mar 21 at 5:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Certainly the OP asked about sequences and I think your answer addresses that issue quite well. I was just a bit curious for myself and also thought that nets would be worth a mention. Anyways +1 for the answer.
  $endgroup$
  – Gary Moon
  Mar 21 at 6:00
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

It's true in a metric space, as a metric space is first countable and we can use countable choice to show it (well within ZFC).

It's not true in general: let $X$ be the reals (or any uncountable set) in the co-countable topology. Any sequence in $X$ that converges to $p$ is equal to $p$ after finitely many terms (eventually constant); many proofs exist for this on this site.

So $0 in overlineA$ where $A= (0,infty)$ but no sequence in $A$ converges to $0$. Also, the identity map from $X$ to $Y$, the reals in the usual topology, is not continuous but trivially sequentially continuous.

share|cite|improve this answer

answered Mar 21 at 5:24

Henno BrandsmaHenno Brandsma

115k349125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I may be mistaken as I'm certainly not a topologist, but can't these statements be generalized in terms of nets?
  $endgroup$
  – Gary Moon
  Mar 21 at 5:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GaryMoon sure, there is always a net converging from $A$ to $p$ and continuity is net-convergence preservation. But the OP asked for sequences.
  $endgroup$
  – Henno Brandsma
  Mar 21 at 5:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Certainly the OP asked about sequences and I think your answer addresses that issue quite well. I was just a bit curious for myself and also thought that nets would be worth a mention. Anyways +1 for the answer.
  $endgroup$
  – Gary Moon
  Mar 21 at 6:00
  
  

  
  
  
  

add a comment | 

2

$begingroup$

It's true in a metric space, as a metric space is first countable and we can use countable choice to show it (well within ZFC).

It's not true in general: let $X$ be the reals (or any uncountable set) in the co-countable topology. Any sequence in $X$ that converges to $p$ is equal to $p$ after finitely many terms (eventually constant); many proofs exist for this on this site.

So $0 in overlineA$ where $A= (0,infty)$ but no sequence in $A$ converges to $0$. Also, the identity map from $X$ to $Y$, the reals in the usual topology, is not continuous but trivially sequentially continuous.

share|cite|improve this answer

answered Mar 21 at 5:24

Henno BrandsmaHenno Brandsma

115k349125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I may be mistaken as I'm certainly not a topologist, but can't these statements be generalized in terms of nets?
  $endgroup$
  – Gary Moon
  Mar 21 at 5:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GaryMoon sure, there is always a net converging from $A$ to $p$ and continuity is net-convergence preservation. But the OP asked for sequences.
  $endgroup$
  – Henno Brandsma
  Mar 21 at 5:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Certainly the OP asked about sequences and I think your answer addresses that issue quite well. I was just a bit curious for myself and also thought that nets would be worth a mention. Anyways +1 for the answer.
  $endgroup$
  – Gary Moon
  Mar 21 at 6:00
  
  

  
  
  
  

add a comment | 

$begingroup$

It's true in a metric space, as a metric space is first countable and we can use countable choice to show it (well within ZFC).

It's not true in general: let $X$ be the reals (or any uncountable set) in the co-countable topology. Any sequence in $X$ that converges to $p$ is equal to $p$ after finitely many terms (eventually constant); many proofs exist for this on this site.

So $0 in overlineA$ where $A= (0,infty)$ but no sequence in $A$ converges to $0$. Also, the identity map from $X$ to $Y$, the reals in the usual topology, is not continuous but trivially sequentially continuous.

share|cite|improve this answer

answered Mar 21 at 5:24

Henno BrandsmaHenno Brandsma

115k349125

$endgroup$

share|cite|improve this answer

answered Mar 21 at 5:24

Henno BrandsmaHenno Brandsma

115k349125

answered Mar 21 at 5:24

Henno BrandsmaHenno Brandsma

115k349125

answered Mar 21 at 5:24

Henno BrandsmaHenno Brandsma

115k349125