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

1. Mar 15, 2014

### wotanub

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: Sep 25, 2014
2. Mar 15, 2014

### 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 $\mathbb{R}^d$.

3. Mar 15, 2014

### micromass

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.

4. Mar 15, 2014

### wotanub

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

5. Mar 15, 2014

### micromass

6. Mar 16, 2014

### Hornbein

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: Sep 25, 2014
7. Apr 9, 2014

### nucl34rgg

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

8. Apr 9, 2014

### HallsofIvy

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. Apr 9, 2014

### micromass

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. Apr 9, 2014

### D H

Staff Emeritus
Godwin's law meets topology!