# boundary point

• April 30th 2010, 12:41 PM
poorna
boundary point
What exactly are boundary points?

Are they just limit points? because i saw an example which said the bounday points of punctured unit open ball ie {(x,y)| x^2 + y^2 <1 , (x,y) not (0,0) } are the points described by x^2 + y^2 =1and the origin.

iThese are the limit points right?

Then what are the boundary points of Q --is it R?
• April 30th 2010, 01:04 PM
Plato
Quote:

Originally Posted by poorna
What exactly are boundary points?
Are they just limit points? NO
Then what are the boundary points of Q --is it R?

In the space of real numbers consider
$A = (0,1] \cup \{ 2\}$ then $\beta(A)=\{0,1,2\}$.
$2$ is not a limit point!
• April 30th 2010, 10:11 PM
poorna
Ah okay. I looked up the definition of a boundary point. It says a point p is a boundary point of A if every neighbourhood around p contains atleast one pointof A and one point not in A.

Then how is {2} the boundary point. A nbd around 2 contains only 2 from A right?

And The set of rationals - Q. I think the set of all boundary points of Q is R, the real line. Am I right?
• May 1st 2010, 05:32 AM
Plato
Quote:

Originally Posted by poorna
Ah okay. I looked up the definition of a boundary point. It says a point p is a boundary point of A if every neighbourhood around p contains atleast one pointof A and one point not in A.

Then how is {2} the boundary point. A nbd around 2 contains only 2 from A right?

And The set of rationals - Q. I think the set of all boundary points of Q is R, the real line. Am I right?

Well first of all $2\in A$.
Now if $\delta > 0$ then the neighbourhood $(2-\delta,2+\delta)$ contains $2+\frac{\delta}{2}$ and $2+\frac{\delta}{2}\notin A$
Thus every neighbourhood of $2$ contains one pointof $A,~2,$ and one point not in $A$.
• May 1st 2010, 06:11 AM
HallsofIvy
I am assuming you are talking about metric topologies.

A point, p, is said to be an "interior" point of set A if there exist $\delta> 0$ such that the set $\{x| d(x,p)< \delta\}$, the " $\delta$ neighborhood of p", is a subset of A.

A point, p, is said to be an "exterior" point of set A if it is an interior point of the complement of A.

A point, p, is said to be a "boundary" point of set A if it is neither an interior point nor an exterior point of A.

Note that these threes sets, the interior points of A, the exterior points of A, and the boundary points of A, "partition" the entire space- every point of the space is in one and only one of those. Note also that A and the complement of A "swap" interior and exterior points- the interior points of A are the exterior points of the complement of A and vice- versa. But A and its complement have the same boundary points.

No, the concepts of "boundary point" and "limit point" are not the same. For example, any point in the interval (0, 1) is a "limit point"- given any p in that interval, there exist a neighborhood of p containing an infinite number of points in (0, 1)- alternatively, given any p in that interval, there exist a sequence of points in the interval that converges to p. But the only boundary points are a and b.

Conversely, consider the set {1/n| n a positive integer}. Every point in that set, together with 0, is a boundary point of the set but 0 is the only limit point.
• May 1st 2010, 08:58 AM
poorna
Oh thanks i think i got it :)

So for The set of rationals - Q. I think the set of all boundary points of Q is R, the real line. Am I right??