I Discrete Topology and Closed Sets

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Summary
proof. Topological space with all closed sets is a discrete space
I am trying to learn some topology and was looking at a problem in the back of the book asking to show that a topological space with the property that all set are closed is a discrete space which, as understand it, means that all possible subsets are in the topology and since all subsets are closed for each set in the topology the compliment must be in as well. Lets say I have a very simple topology on set X
where X={a, b, c, d, e, f} such that the topology is given by

{∅, X, {a, b}, {c, d}, {e, f}, {c, d, e, f}, {a, b, e, f}, {a, b, c , d}}

Unless I am overlooking something, this is a topology with all closed sets, (they are all open as well because this is a topology), that is not a discrete topology because it does not contain singleton sets or am I missing something?
 

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Consider a topological space ##(X, \mathcal{T})##. Such a space is called discrete if ##\mathcal{T}= \mathcal{P}(X) = 2^X##, the power set of ##X## (= all possible subsets of ##X##). In particular, this means that singelton sets must be contained in a discrete topology (conversely, if a topology contains all singeltons, it must be discrete since every set is the union of singeltons and arbitrary unions of open sets remain open).

Note that all sets are open is equivalent with all sets are closed.

The complement of every set in your topology is contained in the topology. This means that all closed sets of your topology are also open (and conversely, every open set is also closed). So yes, you are right. This space is not discrete since the topology you wrote down is not the power set of ##\{a,b,c,d,e,f\}##.

A set that is both open and closed is called clopen. Some more terminology: a topological space is connected if it has no clopen subsets, except ##\emptyset## and ##X##.
 
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So here is the question as written, "Let (X,T) be a topological space with the property that every subset is closed. Prove that it is a discrete space."

The definition of a closed set in the book is given by:

" Let (X,T) be a topological space. A subset S of X is said to be a closed set in (X,T) if its complement, namely X\S, is open in (X,T)"
 

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Take an arbitrary subset ##A \subseteq X##. You have to show that ##A## is open. But this is obvious: you are given that every subset of ##X## is closed, so in particular ##X\setminus A## is closed. By your definition of closed, this means that the complement ##X\setminus (X \setminus A)## is open. But ##X\setminus (X \setminus A) = A## and you are done.
 
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Yeah, but I am not sure how this proves it is a discrete space as the example I gave above is all closed (They are open sets they are referred to in the text as "clopen" by the definition) but it does not contain any singletons. So given that the above example consists of all closed sets that is not a discrete topology because it contains no singletons, thus it cannot be proven?
 

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Yeah, but I am not sure how this proves it is a discrete space as the example I gave above is all closed (They are open sets they are referred to in the text as "clopen" by the definition) but it does not contain any singletons. So given that the above example consists of all closed sets that is not a discrete topology because it contains no singletons, thus it cannot be proven?
What is your definition of discrete space?
 
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"Let X be any non-empty set and let T be the collection of all subsets of X. Then T is called the discrete topology on the set X . The topological space (X,T ) is called a discrete space"

Then is goes on to show

"If (X,T ) is a topological space such that for every x ∈ X, the singleton set {x} is in T, then T is the discrete topology"
 

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"Let X be any non-empty set and let T be the collection of all subsets of X. Then T is called the discrete topology on the set X . The topological space (X,T ) is called a discrete space"

Then is goes on to show

"If (X,T ) is a topological space such that for every x ∈ X, the singleton set {x} is in T, then T is the discrete topology"
And why are you confused then? In your example ##X = \{a,b,c,d,e,f\}## and the topology you wrote down does not contain singeltons, so it isn't a discrete space.

In your example, it just happens that every open set is also closed, but this does not mean in any way that the space is discrete.
 
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So the actual problem is

"Let (X,T) be a topological space with the property that every subset is closed. Prove that it is a discrete space."

So the example is just to show that I can create an arbitrary topology with all closed sets that is not a discrete space. Thus, I have come up with a counter example which proves that a topology with all closed sets is not necessarily a discrete space unless I am missing something.

BTW thanks for your help, proofs are not my strong point, as I mostly work with applied math, but I am trying to get better at them.
 

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So the actual problem is

"Let (X,T) be a topological space with the property that every subset is closed. Prove that it is a discrete space."

So the example is just to show that I can create an arbitrary topology with all closed sets that is not a discrete space. Thus, I have come up with a counter example which proves that a topology with all closed sets is not necessarily a discrete space unless I am missing something.

BTW thanks for your help, proofs are not my strong point, as I mostly work with applied math, but I am trying to get better at them.
I see what the problem is. You should understand the sentence "every subset is closed" as "every subset of X is closed" and NOT "every subset in the topology is closed". Hope this clarifies the issue.
 
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I think so but the proof then seems a little trivial

By definition if the topology has every set of X it is a discrete topology. Since it is a discrete topology it must include all compliments in the topology and thus every set
 
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I think so but the proof then seems a little trivial

By definition if the topology has every set of X it is a discrete topology. Since it is a discrete topology the compliment to each subset must be in the topology and therefore every set is closed.

Anyways thanks for your help
 

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I think so but the proof then seems a little trivial

By definition if the topology has every set of X it is a discrete topology. Since it is a discrete topology the compliment to each subset must be in the topology and therefore every set is closed.

Anyways thanks for your help
The proof is trivial. These are just exercises to see that you understood the definitions. Things will get harder :)
 

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