The a set is open iff its complement is closed?

[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?

 

[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 $\mathbb{R}^d$.

 

[micromass]

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.

 

[wotanub]

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

 

[micromass]

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

 

[Hornbein]

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.

 

[nucl34rgg]

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

 

[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.

 

[micromass]

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]

https://www.youtube.com/watch?v=SyD4p8_y8Kw

Godwin's law meets topology!