S=(-inf,1]U{3/2,5/3,7/4,...,(2n-1)/n,...}
Question:
Give the boundary points, the interior points, the accumulation points, the isolated points. Give sup S and inf S. And is S compact?
Please help!
Thanks.
S=(-inf,1]U{3/2,5/3,7/4,...,(2n-1)/n,...}
Question:
Give the boundary points, the interior points, the accumulation points, the isolated points. Give sup S and inf S. And is S compact?
Please help!
Thanks.
The set of interior points is easy. When you are looking for accumulation points, recall that given any real number, there is a sequence of rational numbers that converges to it.
But exactly what kind of "help" do you want? Do you know the definitions of any of those things?
@CountingPenguins; I want to reinforce the above. I do not want to think that you are just fishing hoping to get some free answers to questions that you need to answer for yourself.
To prove that wrong, please post what you know about the definitions of:
interior, boundary, and accumulation points.
Yes that is the correct set. No I was not fishing, but I did not include my answers for ease of reading. But I will include them here.
interior points: (-$\displaystyle \infty$,1)U{$\displaystyle \frac{3}{2}$,$\displaystyle \frac{5}{3}$,...,(2n-1)/n,....,2}
boundary points: {1,$\displaystyle \frac{3}{2}$,2}
accumulation points: {1,$\displaystyle \frac{3}{2}$,2}
isolated points: ?
sup S: 2
inf S: does not exist
Is S compact? no because there are numbers between 3/2 and 5/3 that are not in the set.
I apologize for the confusion.
Any assistance in reviewing my work is appreciated.
You did do what you were asked to do.
You did not provide us with working definitions.
You got most all of the above wrong. But we cannot help if we do not know your definitions.
You are correct that $\displaystyle S$ not compact, but the reason you gave has absolutely nothing to do with compactness. What is compactness?
Interior points are points not in the boundary. Everything I wrote except 2.
A boundary point intersects S and its complement. So this would be 2.
x is an accumulation point if every deleted neighborhood of x contains a point of S. 2
isolated point if x is an element of S and is not an accumulation point of S. everything except 2.
Are you kidding us? Those are not proper definitions. Only the one for accumulation is even close to correct.
Where did you get these? What text material are you using. Please give us the exact wording of the definitions.
I will show how absurd two of them are.
You say “A boundary point intersects S and its complement”.
Well it is literally impossible for a point to intersect a set and its complement.
Second, you say that “Interior points are points not in the boundary.”
That means that exterior points are boundary points that is absurd.
So if those are the exact wordings, then I will give you the answers if you name the textbook or the school and lecturer that you got these definitions from. They need to be reported as frauds.
The fault is mine.
Steven Lay, Analysis
A point x is an isolated point of S if x is an element of S and x is not an accumulation point of S.
A point x in $\displaystyle \Re$ is an interior point of a set S if there exists a neighborhood N of x such that N is a proper subset of S.
A point x in $\displaystyle \Re$ is a boundary point of a set S if every neighborhood of x intersects both S and the complement of S in $\displaystyle \Re$.
A point x is an accumulation point of a set S if every deleted neighborhood of x contains a point of S.
The interior is $\displaystyle \left( { - \infty ,1} \right) $
The boundary is $\displaystyle \left\{ {\frac{{2n - 1}}{n}:n =1, 2,3, \cdots }\right\}$.
The set of accumulation points is $\displaystyle \left( { - \infty ,1} \right]$.
A boundary point which is not an accumulation point is isolated.
Thus 1 is not an isolated point.