# The a set is open iff its complement is closed?

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

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

Last edited by a moderator:

pwsnafu
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##.

1 person
What am I missing here? Does it have to do with choosing the complement in $S$ rather than the complement in $\mathbb{R}^d$?

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.

1 person
I see. Thanks for the speedy replies... The names "closed" and "open" are really unfortunate it seems.

I see. Thanks for the speedy replies... The names "closed" and "open" are really unfortunate it seems.

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

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

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:
Yes, the empty set and the whole space are clopen.

HallsofIvy
Homework Helper
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.

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

D H
Staff Emeritus