Closed Subsets in a Toplogical space ....

[Math Amateur]

TL;DR Summary

I have a very basic question about whether closed subsets of a topological space $(X, \tau)$ that are not clopen belong to the space ... I suspect that they do not ...

I am reading Sasho Kalajdzievski's book: "An Illustrated Introduction to Topology and Homotopy" and am currently focused on Chapter 3: Topological Spaces: Definitions and Examples ... ...

I need some help in order to fully understand Kalajdzievski's definition of a closed set in a topological space ...The relevant text reads as follows:

As I understand it many closed subsets of the underlying set $X$ of a topological space $(X, \tau)$ do not belong to the topological space because they are not open ... i.e. they are not clopen sets ...

Is my interpretation of the above situation correct ... ... ?Help will be appreciated ...

Peter

==================================================================================It may help readers of the above post to have available Kalajdzievski's definition of a topological space ... so I am providing the same ... as follows:

Hope that helps ...

Peter

 

[suremarc]

Yes, that is correct. $\tau$ is the collection of open sets, so saying a set $U$ is in $\tau$ means exactly that $U$ is open.

 

[martinbn]

Just point out that it is not clear what you are asking. A subset, which is not open, doesn't belong to the topology of the psace, but in a sense it belongs to the space, after all it is a subset.

 

[mathwonk]

Peter, I think you are confusing the two statements: "a set U belongs to [is a subset of] the topological space X" and "a set U belongs to the topology defining the topological space X". there are many sub sets of a topological space, some open, some closed, some both, some neither. the ones that are open are the ones that belong to "the topology". i.e. a subset U of X is open if and only if U belongs to "the topology of X". "The topology" of X is the collectionmof those subsets that are open in X. You are advised to study this for the special case of metric spaces.

 

[Math Amateur]

Thanks to suremarc and martinbn ... appreciate your help ...

Particular thank you to mathwonk ... yes, I was confused in exactly that way ... thanks so much for clarifying the issue ...

Peter