The closure of an open set A, strictly bigger than A itself?

[adam512]

Hi there!

Is the following true?

Suppose A is an open set and not closed. Cl(A) is closed and contains A, hence it contains at least one point not in A.

If A is both open and closed it obviously does not hold.

 

[CompuChip]

In the axioms for a topology, we assume the whole space is an open set. Its closure wouldn't be larger.

But TS said

Suppose A is an open set and not closed.

and the whole space is closed (because it's the complement of the empty set).

As you've worded it, your statement is true. Suppose that cl(A) is not strictly bigger. By definition, it is not smaller either (it's the smallest closed set that contains A), so it must be exactly A.

 

[Stephen Tashi]

But TS said

and the whole space is closed (because it's the complement of the empty set).

As you've worded it, your statement is true. Suppose that cl(A) is not strictly bigger. By definition, it is not smaller either (it's the smallest closed set that contains A), so it must be exactly A.

I agree. I realized the mistake and deleted my post, before yours appeared.

 

[HallsofIvy]

They just won't let you pretend you didn't make a mistake!

(They do that to me all the time!)

 

[CompuChip]

Luckily you hardly make any Halls :P

Sorry Stephen, sometimes I leave the tab open and reply a bit later without hitting refresh first.

 

[mathwonk]

adam512, there is an interesting concept related to your question, namely connectedness.

a space is connected if and only if the only sets that are both open and closed are the empty set and the whole space.

In other words, in a connected space, no "proper" open subset can be also closed.

Thus in a connected space, the closure of a non empty proper open subset A, will be strictly larger than A.