# Connected sets in a topological space

1. May 28, 2009

### logarithmic

The definition of Y being connected in a topological space (X, tau) is that you can't find two non-empty, open and disjoint sets whose union is Y.

This doesn't quite make much intuitive sense to me.

For example, consider R with the usual topology. Then clearly, Y= [0,1] union [2,3] is not connected. That means you CAN find two non-empty, open and disjoint, sets whose union is Y. But what are they?

I can't seem to think of 2 open sets in R whose union is [0,1] union [2,3].

2. May 28, 2009

### slider142

The set [0,1] is closed in R, but open in the relative topology of Y (with respect to the usual topology on R). The same is true of [2,3].

3. May 28, 2009

### HallsofIvy

The problem is that your definition is wrong.

A topological space is connected if and only if you cannot find two such open sets. That is not the case for a subset of a topological space. (You can also replace "open" with "closed".)

For a subset, A, of a topological space, X, A is connected if and only if it is not the union of two separated sets.

Sets, U and V, are said to be separated if and only if $\overline{U}\cap V= \phi$ and $U\cap \overline{V}= \phi$, where $\overline{U}$ is its closure.

In your example, we can take U= [0, 1] and V= [2, 3].

Of course, we can always think of a subset of a topological space as being a topological space with the relative topology: A subset of A is open "in A" if and only if it is the intersection of A with a set open in X.

If $A= [0, 1]\cup [2, 3]$, then, since $[0, 1]= A\cup (-1/2, 3/2)$ and $[2, 3]= A\cap (3/2, 7/2)$ both [0, 1] and [2, 3] are open in A (or "open relative to A").

Notice that, in A (or "relative to A") [0, 1] is the complement of [2, 3] and [2, 3] is open, so [0, 1] is also closed. Similarly [2, 3] is also both open and closed in A. It is always true that "connectedness components" of a space (connected sets not properly contained in any connected set) are both open and closed.

Last edited by a moderator: May 28, 2009
4. May 28, 2009

### logarithmic

Thanks for clearing that up. I'm not sure what your definition of separated sets is meant to say though, I think you made a typo there.

5. May 28, 2009

### HallsofIvy

You are right. For some reason I used \overbar(U) instead of \overline in my LaTex to denote "closure" and it didn't work! I have editted my original post.

Last edited by a moderator: May 28, 2009