Confused with closure and interior

[Damascus Road]

Greetings all,

I'm looking at some examples in the Topology: Pure and Applied text.

Looking at example 2.1 Consider A=[0,1) as a subset of R with the standard topology. Then Aint=(0,1) and Aclos=[0,1].

Can someone explain to me why the union of all open sets in A is that?

Furthermore, example 2.3

Consider A=[0,1) as a subset of R in the finite complement topology. Here Aint = 0 because there are no nonempty open sets contained in [0,1). Since A is infinite, and the only infinite closed set in this topology is R, it follows that Aclos=R.

I have the same question as above, but also,why is A infinite? And more importantly, the book described R as being open when proving the the finite complement is a topology, and now they are saying its closed?

Please help!

 

[slider142]

Greetings all,

I'm looking at some examples in the Topology: Pure and Applied text.

Looking at example 2.1 Consider A=[0,1) as a subset of R with the standard topology. Then Aint=(0,1) and Aclos=[0,1].

Can someone explain to me why the union of all open sets in A is that?

Is which? The union of all open sets in A, (0, 1), is the interior of A. It is trivial that there is no larger open subset of A, as A - (0, 1) is a singleton. Constructively, take the union of all disc neighborhoods of points in A whose intersection with R - A is null. The closure of A is the set of all points for which every neighborhood has an intersection with A that is not the null set.

Furthermore, example 2.3

Consider A=[0,1) as a subset of R in the finite complement topology. Here Aint = 0 because there are no nonempty open sets contained in [0,1). Since A is infinite, and the only infinite closed set in this topology is R, it follows that Aclos=R.

I have the same question as above, but also,why is A infinite? And more importantly, the book described R as being open when proving the the finite complement is a topology, and now they are saying its closed?

The entire space is always both open and closed in any topology. This is a consequence of the definition of open and closed subsets of R in combination with the null set being a subset of every set, and is stated explicitly in the definition of a topological space.
The finite complement topology uses the amount of elements of a set to define open or closed; there are an infinite amount of elements in any real interval, which is why A is described as being infinite, and therefore open.

 

[Damascus Road]

Thank you for the reply, I had forgotten about that property of being both open and closed.

You're explanation has helped a lot!

The thing I am still struggling with is the closure now, the "intersection of all closed sets in A".

First, could I not choose a "set" that is closed in A that doesn't include 0 or 1?
Then the intersection would be empty...

Second, if the set A is open on 1 (i.e. doesn't reach one) how can it be included in the closure?

 

[disregardthat]

The thing I am still struggling with is the closure now, the "intersection of all closed sets in A".

The closure is the intersection of all closed sets containing A. So the closure will always contain A. You can argue that

1) [0,1] is a closed set containing A.
2) 1 is a limit point of A, hence must be an element of the closure.

 

[slider142]

The thing I am still struggling with is the closure now, the "intersection of all closed sets in A".

First, could I not choose a "set" that is closed in A that doesn't include 0 or 1?
Then the intersection would be empty...

You can't choose sets in any manner, you must intersect all closed sets that contain A.

Second, if the set A is open on 1 (i.e. doesn't reach one) how can it be included in the closure?

A closed set C containing A must contain 1. If it did not contain 1, then the boundary of the closed set must necessarily include a number r less than 1. Under the standard topology, you can simply choose a disc neighborhood around that number with radius 1-r that includes points not in C, but in A, contradicting the requirement that C contains A.