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On accumulation points of sequences

Accumulation points of sequences as limits of subsequences?Divergent Complex sequence with distinct points and without accumulation point tends to $infty$?Accumulation point is a limit of some sequences in first-countable space with Axiom of Choice.How to show $[0, fracpi2) cup (frac3pi4, 2pi)$ is sequentially separated?Proof verification: Set of partial limits of a sequence is closedIs every accumulation point of the set of values of sequence $x_n$ also the accumulation point of the sequence $x_n$.Does every iterative definition (or at least the one describe in the details) require the use of the Axiom of Choice?Any subset $X$ of $mathbbR^n$ satisfying property C is compact.A convergent sequence has precisely one accumulation pointEquivalent statements about accumulation point of a sequence?

0

$begingroup$

Theorem $1$. Statements below are equivalent:

($1$) $x$ is accumulation point of sequence $(x_n)$

($2$) In every neighborhood of $x$ there are infinitely many elements of sequence

Theorem $2$. If $x notin$ $x_n: n in N$, statements below are equivalent:

($1$) $x$ is accumulation point of sequence $(x_n)$

($2'$) In every neighborhood of $x$ there is at least one element of sequence

I understand that when $x notin$ $x_n: n in N$, we can replace ($2$) with ($2'$) and vice versa, but I can't see the difference when $x in$ $x_n: n in N$, too. So, is there really a difference ?

sequences-and-series

share|cite|improve this question

asked Mar 17 at 13:58

ubermenschubermensch

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If x belongs to the sequence, there always been one element of that sequence in the neighborhood of x (itself)
  $endgroup$
  – vladz
  Mar 17 at 14:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And if we say that there is at least one element of the sequence different from $x$ ?
  $endgroup$
  – ubermensch
  Mar 17 at 14:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then you are good. If you have at least one in any neighborhood, you have one in any sub-neighborhood, thus have infinitely many in any neighborhood
  $endgroup$
  – vladz
  Mar 17 at 14:58
  

  
  
  
  

add a comment | 

0

$begingroup$

Theorem $1$. Statements below are equivalent:

($1$) $x$ is accumulation point of sequence $(x_n)$

($2$) In every neighborhood of $x$ there are infinitely many elements of sequence

Theorem $2$. If $x notin$ $x_n: n in N$, statements below are equivalent:

($1$) $x$ is accumulation point of sequence $(x_n)$

($2'$) In every neighborhood of $x$ there is at least one element of sequence

I understand that when $x notin$ $x_n: n in N$, we can replace ($2$) with ($2'$) and vice versa, but I can't see the difference when $x in$ $x_n: n in N$, too. So, is there really a difference ?

sequences-and-series

share|cite|improve this question

asked Mar 17 at 13:58

ubermenschubermensch

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If x belongs to the sequence, there always been one element of that sequence in the neighborhood of x (itself)
  $endgroup$
  – vladz
  Mar 17 at 14:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And if we say that there is at least one element of the sequence different from $x$ ?
  $endgroup$
  – ubermensch
  Mar 17 at 14:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Then you are good. If you have at least one in any neighborhood, you have one in any sub-neighborhood, thus have infinitely many in any neighborhood
  $endgroup$
  – vladz
  Mar 17 at 14:58
  

  
  
  
  

add a comment | 

$begingroup$

Theorem $1$. Statements below are equivalent:

($1$) $x$ is accumulation point of sequence $(x_n)$

($2$) In every neighborhood of $x$ there are infinitely many elements of sequence

Theorem $2$. If $x notin$ $x_n: n in N$, statements below are equivalent:

($1$) $x$ is accumulation point of sequence $(x_n)$

($2'$) In every neighborhood of $x$ there is at least one element of sequence

I understand that when $x notin$ $x_n: n in N$, we can replace ($2$) with ($2'$) and vice versa, but I can't see the difference when $x in$ $x_n: n in N$, too. So, is there really a difference ?

sequences-and-series

share|cite|improve this question

asked Mar 17 at 13:58

ubermenschubermensch

1

$endgroup$

share|cite|improve this question

asked Mar 17 at 13:58

ubermenschubermensch

1

share|cite|improve this question

asked Mar 17 at 13:58

ubermenschubermensch

1

asked Mar 17 at 13:58

ubermenschubermensch

1

asked Mar 17 at 13:58

ubermenschubermensch

1