Why Does B Include the Segment [a,c) in Topological Space X?

[qinglong.1397]

I am reading H. Croom's Principles of Topology and in page 139, he gave an example 5.2.5 to show that two points in the same member of any separation of the topological space X, might not belong to the same component of X.

In this example, the space X is a subspace of 2 dimensional Euclidean space with the usual topology. X consists of a sequence of line segments converging to a line segment whose midpoint c has been removed. Then [a,c) is one component of X, but why?

You see, since the sequence of the segments converge to [a,b], any open set containing [a,c) must intersect with infinitely many segments in the sequence. Therefore, [a,c) must belong to some "bigger" subset of X, which is a component. But why is my reasoning wrong?

Thank you!

 

[Office_Shredder]

Suppose that $A = [a,c)\cup B$ is a connected component for some B. Let $x \in B$ be a point which is on the nth line down in the picture (where the 1st line down is the top line drawn, the second line down is the second from the top, etc.). Then we can split B into two pieces,

$$B = C \cup D$$
where C is everything in B in the top n lines, and D is everything below. It's clear that C is non-empty as it contains the point x that we had in B by construction, and D contains [a,c) so is non-empty as well. But D and C can be shown to be disjoint open sets.

Alternatively it is very obvious that each discrete line except for the bottom one is a connected component of X.The trick here is that contrary to intuition, the connected components of a topological space do not have to be open sets, they are in fact closed sets. If there are finitely many then they will be open but if you have infinitely many connected components they won't all be except in special cases.

 

[qinglong.1397]

Suppose that $A = [a,c)\cup B$ is a connected component for some B. Let $x \in B$ be a point which is on the nth line down in the picture (where the 1st line down is the top line drawn, the second line down is the second from the top, etc.). Then we can split B into two pieces,

$$B = C \cup D$$
where C is everything in B in the top n lines, and D is everything below. It's clear that C is non-empty as it contains the point x that we had in B by construction, and D contains [a,c) so is non-empty as well. But D and C can be shown to be disjoint open sets.

Alternatively it is very obvious that each discrete line except for the bottom one is a connected component of X.


The trick here is that contrary to intuition, the connected components of a topological space do not have to be open sets, they are in fact closed sets. If there are finitely many then they will be open but if you have infinitely many connected components they won't all be except in special cases.

Thanks! But why does B contain [a,c)?