The a set is open iff its complement is closed?

  • Thread starter wotanub
  • Start date
  • #1
230
8

Around the 4 minute mark the lecturer makes this statement, but I am not convinced this is true. I accept that

(1) if a set is closed, its complement is open.

but consider the converse.

Consider an open ball [itex]S[/itex] of some arbitrary radius centered at the origin (in whatever dimension [itex]d[/itex] you want). [itex]S[/itex] is trivially subset of itself, and its complement (in [itex]S[/itex] that is) is [itex]S^c = S \setminus S[/itex] is the empty set [itex]ø[/itex]. But the empty set is open, implying [itex]S[/itex] is closed by (1), and this is a contradiction since we started with choosing [itex]S[/itex] as open.

What am I missing here? Does it have to do with choosing the complement in [itex]S[/itex] rather than the complement in [itex]\mathbb{R}^d[/itex]?
 
Last edited by a moderator:

Answers and Replies

  • #2
pwsnafu
Science Advisor
1,080
85
If you take ##S## as your entire space (which is what you have done), then ##S## is by definition both open and closed in itself. ##S## is not closed relative to the entire ##\mathbb{R}^d##.
 
  • Like
Likes 1 person
  • #3
22,129
3,297
What am I missing here? Does it have to do with choosing the complement in [itex]S[/itex] rather than the complement in [itex]\mathbb{R}^d[/itex]?

Yes. Here, you regard ##S## as a space in itself. You didn't regard ##S## as the subset of some bigger space.

In this case, ##S## is indeed open and closed in ##S## (we call that clopen). However, ##S## is open in ##\mathbb{R}^d##, but not closed in ##\mathbb{R}^d##.

So the difference is the big space you're working in:
If you're working in ##S##, Then closed in ##S## means that the complement relative to ##S## is open. So ##S\setminus S## is open, which is true.
If you're working in ##\mathbb{R}^d##, then closed in ##S## means that the complement relative to ##\mathbb{R}^d## is open. So ##\mathbb{R}^d\setminus S## is open, which is true.

So open and closed are relative notions, depending on the bigger space.
 
  • Like
Likes 1 person
  • #4
230
8
I see. Thanks for the speedy replies... The names "closed" and "open" are really unfortunate it seems.
 
  • #5
22,129
3,297
I see. Thanks for the speedy replies... The names "closed" and "open" are really unfortunate it seems.

https://www.youtube.com/watch?v=SyD4p8_y8Kw
 
  • #6
699
337

Around the 4 minute mark the lecturer makes this statement, but I am not convinced this is true. I accept that

(1) if a set is closed, its complement is open.

but consider the converse.

Consider an open ball [itex]S[/itex] of some arbitrary radius centered at the origin (in whatever dimension [itex]d[/itex] you want). [itex]S[/itex] is trivially subset of itself, and its complement (in [itex]S[/itex] that is) is [itex]S^c = S \setminus S[/itex] is the empty set [itex]ø[/itex]. But the empty set is open, implying [itex]S[/itex] is closed by (1), and this is a contradiction since we started with choosing [itex]S[/itex] as open.

What am I missing here? Does it have to do with choosing the complement in [itex]S[/itex] rather than the complement in [itex]\mathbb{R}^d[/itex]?

I've been away from all this for a long time, but seem to recall that the empty set is vacuously both open and closed.
 
Last edited by a moderator:
  • #7
148
14
Yes, the empty set and the whole space are clopen.
 
  • #8
HallsofIvy
Science Advisor
Homework Helper
41,847
964
In fact, if a set is not "connected" then all "connectedness components" (connected set which are not properly contained in any connected sets) are both open and closed.
 
  • #9
22,129
3,297
In fact, if a set is not "connected" then all "connectedness components" (connected set which are not properly contained in any connected sets) are both open and closed.

Closed yes. Open no, consider ##\mathbb{Q}##. To have the connected components open you need the notion of locally connectedness: http://en.wikipedia.org/wiki/Locally_connected_space
 
  • #10
D H
Staff Emeritus
Science Advisor
Insights Author
15,415
686
https://www.youtube.com/watch?v=SyD4p8_y8Kw
Godwin's law meets topology!
 

Related Threads on The a set is open iff its complement is closed?

Replies
7
Views
1K
Replies
8
Views
79
Replies
4
Views
6K
Replies
14
Views
7K
Replies
2
Views
1K
  • Last Post
Replies
14
Views
3K
Replies
6
Views
10K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
2
Views
3K
Top