Accumulation Points in [0,1]: Understanding the Boundaries

[mbarby]

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...

 

[micromass]

Hi mbarby!

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].

 

[mbarby]

what really confuses me here is that:
lets assume 1 is accumulation point in (0,1)
then shouldn't it be contained in an interval like (1-e, 1+e) (e=epsilon)
but we don't have 1+e since it exceeds interval border..
where am i wrong now ?

 

[micromass]

what really confuses me here is that:
lets assume 1 is accumulation point in (0,1)
then shouldn't it be contained in an interval like (1-e, 1+e) (e=epsilon)
but we don't have 1+e since it exceeds interval border..
where am i wrong now ?

You are considering (0,1) as a subset of the space $\mathbb{R}$, 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?

 

[mbarby]

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 can't 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 isn't there a rule against it ? etc.
thx for the quick explanations by the way.

 

[micromass]

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 can't 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 isn't 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).

 

[mbarby]

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 ?

 

[SteveL27]

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.

 

[micromass]

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...

 

[mbarby]

thx all guys,
that eased my mind. i was tearing myself apart to understand where i was making the mistakes :/ ...