Question about isolated points.

  • Thread starter Thread starter cragar
  • Start date Start date
  • Tags Tags
    Points
Click For Summary

Homework Help Overview

The discussion revolves around the properties of a set containing the single element \(\pi\) on the real line, specifically questioning whether this set is closed and the nature of isolated points.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definition of closed sets and limit points, with some asserting that the set is closed due to containing its only point. Others question the implications of the definition of limit points and express a desire for further clarification on the original poster's doubts.

Discussion Status

The discussion is ongoing, with participants offering differing perspectives on the nature of the set. Some guidance has been provided regarding the definitions involved, and there is an exploration of assumptions about the set being closed.

Contextual Notes

There is a noted uncertainty regarding the definitions of limit points and closed sets, as well as the implications of considering the set as closed or not.

cragar
Messages
2,546
Reaction score
3

Homework Statement


If I just had the set containing [itex]\pi[/itex] on the real line.
So this is an isolated point. Is this set closed?

The Attempt at a Solution


I think this set is closed because it contains its limit points, because it only has one point.
Am i thinking about this correctly?
 
Physics news on Phys.org
Hmm, that's an interesting question. I guess it would be closed because that's the only point in the set, and there's no option for an e-neighbourhood around it.
 
I think you are thinking about it correctly but that you are even asking this question makes me wonder. Can you explain your doubts further? What's the exact definition of a limit point?
 
thats what I thought too
 
If we wanted to put things on more definite footing, we could assume that it isn't closed. Then there is some point x1 such that for all E > 0 there is some x in our set such that d(x,x1) < E. However, since x = π in all cases, then the only such point is π, a contradiction.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 21 ·
Replies
21
Views
4K
Replies
4
Views
1K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K