Accumulation point of [0,1]

  • Thread starter mbarby
  • Start date
  • #1
12
0
are 0,1 accumulation points in (0,1) ? how about in [0,1] ?
if 0 and 1 are accumulation points in [0,1] interval what is the open subset they are in ?
i need explanation about this.....pls...
 

Answers and Replies

  • #2
22,089
3,285
Hi mbarby! :smile:

An accumulation point of a set S is a point x such that every neighborhood of x contains a point of S (that is not x). Right?

Well, take (0,1), then every neighborhood around 1 contains a point of (0,1), thus 1 is an accumulation point of (0,1). And so is 0.
Likewise, 0 are 1 are accumulation points of [0,1].
 
  • #3
12
0
what really confuses me here is that:
lets assume 1 is accumulation point in (0,1)
then shouldnt it be contained in an interval like (1-e, 1+e) (e=epsilon)
but we dont have 1+e since it exceeds interval border..
where am i wrong now ?
 
  • #4
22,089
3,285
what really confuses me here is that:
lets assume 1 is accumulation point in (0,1)
then shouldnt it be contained in an interval like (1-e, 1+e) (e=epsilon)
but we dont have 1+e since it exceeds interval border..
where am i wrong now ?
You are considering (0,1) as a subset of the space [itex]\mathbb{R}[/itex], or are you considering (0,1) as a subset of itself?

Anyway, 1 is an accumulation point because every interval (1-e,1+e) around 1 contains a point of (0,1). I don't see what 1+e exceeding the interval boundary has to do with this?
 
  • #5
12
0
sorry for the questions , i am not a math guy. but topology is one of the topics i want to learn. so somethings are as clear to me as it is to you guys.
(0,1) is the space's itself. so we cant take an interval of (1-e, 1+e) without exceeding 1 by +e.
but what i get from your reaction is that we can take the (1-e, 1+e) interval even if it exceeds the boundary, is that right ?
if so why is that possible ? or isnt there a rule against it ? etc.
thx for the quick explanations by the way.
 
  • #6
22,089
3,285
sorry for the questions , i am not a math guy. but topology is one of the topics i want to learn. so somethings are as clear to me as it is to you guys.
(0,1) is the space's itself. so we cant take an interval of (1-e, 1+e) without exceeding 1 by +e.
but what i get from your reaction is that we can take the (1-e, 1+e) interval even if it exceeds the boundary, is that right ?
if so why is that possible ? or isnt there a rule against it ? etc.
thx for the quick explanations by the way.
Ah, I think I get it. But if your space is (0,1), then 1 doesn't belong to this space, thus cannot be an accumulation point. In [0,1], 1 is an accumulation point of (0,1). Indeed, the set (1-e,1] is open in [0,1] and contains points from (0,1).
 
  • #7
12
0
thx a lot this explains a great to me.
is it true , then , if i say any interval having the border of the interval is open, or sth similar to that ?
 
  • #8
795
7
what really confuses me here is that:
lets assume 1 is accumulation point in (0,1)
That's the problem right there. 1 is an accumulation point OF (0,1). But it's wrong to say that 1 is IN (0,1). I believe your use of "in" rather than "of" is causing you to mis-think about this.
 
  • #9
22,089
3,285
thx a lot this explains a great to me.
is it true , then , if i say any interval having the border of the interval is open, or sth similar to that ?
Well, yes, something similar. Things like ]a,1] and [0,a[ are open in [0,1]. But things like [0.5,1] isn't...
 
  • #10
12
0
thx all guys,
that eased my mind. i was tearing myself apart to understand where i was making the mistakes :/ ...
 

Related Threads on Accumulation point of [0,1]

Replies
11
Views
1K
Replies
4
Views
11K
  • Last Post
Replies
3
Views
815
  • Last Post
2
Replies
35
Views
29K
Replies
5
Views
2K
  • Last Post
Replies
6
Views
1K
Replies
3
Views
4K
Replies
4
Views
923
Replies
7
Views
3K
Replies
1
Views
1K
Top