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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?

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?