The a set is open iff its complement is closed?

• wotanub
In summary, the conversation discusses the concept of open and closed sets in terms of their complements and how they relate to each other. The conversation delves into the idea of a set being both open and closed in itself, but not necessarily in a larger space. The speaker explains the difference between looking at a set as a space in itself versus a subset of a bigger space and how that affects the definitions of open and closed. The conversation also touches on the concept of connectedness and how it relates to open and closed sets.
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 $\mathbb{R}^d$?

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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
wotanub said:
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.

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

wotanub said:

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.

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

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.

HallsofIvy said:
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

micromass said:
Godwin's law meets topology!

1. What does it mean for a set to be open?

In mathematics, a set is considered open if all of its points are contained within the set's interior. This means that no points on the boundary or edge of the set are included.

2. How do you determine if a set is open?

A set is open if it meets the definition of being open, which is that all of its points are contained within the set's interior. This can also be determined by checking if the set's complement is closed.

3. What is the complement of a set?

The complement of a set is the set of all elements that are not contained within the original set. In other words, it is the set of elements that are outside of the set.

4. Why is the statement "a set is open iff its complement is closed" important?

This statement is important because it is a fundamental concept in topology and is used to define open and closed sets. It also helps us understand the relationship between open and closed sets and how they complement each other.

5. Can a set be both open and closed?

No, a set cannot be both open and closed. A set is open if all of its points are contained within the interior, while a set is closed if it contains all of its boundary points. Therefore, a set cannot have both interior and boundary points.

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