Solve Connected Set Closure: Show Not Disconnected

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In summary, the conversation discusses a problem of showing that the closure of a connected set is also connected. The suggested approach is to show that if the closure is contained in two disjoint and open sets, then it must be contained in one of them. The conversation then poses the question of solving the problem with one or two additional points and mentions a version of connectedness that makes the problem trivial.
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metalbec
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I'm struggling with something that I suspect is very basic. How do I should that the closure of a connected set is connected? I think I need to somehow show that it is not disconnected, but that's where I'm stuck.

Thanks
 
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  • #2
Call your space [tex]M[/tex]. You want to show that if [tex]\mathrm{cl}(M) \subset X\cup Y[/tex], with [tex]X, Y[/tex] disjoint and open then [tex]\mathrm{cl}(M)[/tex] is contained in either [tex]X[/tex] or [tex]Y[/tex].

Can you go from here?
 
  • #3
Can you solve the problem if the closure has one additional point?
How about two additional points?
 
  • #4
the following version of connectedness makes all possible problems trivial:

a set is connected iff all continuous maps to the set {0,1} are constant.
 

FAQ: Solve Connected Set Closure: Show Not Disconnected

What is the concept of connected set closure?

The connected set closure is a mathematical concept that refers to the smallest closed set that contains a given connected set. It is also known as the closure of a connected set or the topological closure.

What does it mean for a set to be disconnected?

A set is considered disconnected if it can be divided into two non-empty subsets that are separated from each other by a gap or a hole. In other words, there is no continuous path that connects the two subsets.

How can I show that a set is not disconnected?

To show that a set is not disconnected, you can prove that there is no gap or hole that separates the set into two non-empty subsets. This can be done by finding a continuous path that connects any two points in the set, therefore showing that the set is connected.

How does the concept of connected set closure relate to disconnected sets?

The connected set closure is used to determine whether a set is disconnected. If a set is not disconnected, then its connected set closure will be the same as the set itself. However, if a set is disconnected, its connected set closure will be larger than the set, as it includes all the points of the set plus the points that bridge the gap between the disconnected subsets.

Can you provide an example of solving connected set closure to show a set is not disconnected?

Sure, let's take the set of real numbers between 0 and 1, denoted as [0,1]. To show that this set is not disconnected, we can prove that there is a continuous path that connects any two points in the set. For example, we can draw a straight line from 0 to 1, passing through all the points in between, thus showing that the set is connected. Therefore, the connected set closure of [0,1] would be the same as the set itself.

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