# Thread: Connectedness of subsets of R^n

1. ## Connectedness of subsets of R^n

Is there a simple test for connectedness for subsets of R^n? For example, to test if such subsets are compact, you can just check if they are closed and bounded in R^n.

I was wondering if there was such a test for connectedness, as I'm not sure how to decide whether arbitrary sets in R^n are connected. For example, (x,y) in R^2 where at least one of x and y is rational.

Any help would be greatly appreciated

2. A set is connected if and only if it cannot be the union of two non-enpty separarted sets.
That is two set neither of which contains a point nor a limit point of the other.

3. Originally Posted by Magus01
Is there a simple test for connectedness for subsets of R^n? For example, to test if such subsets are compact, you can just check if they are closed and bounded in R^n.

I was wondering if there was such a test for connectedness, as I'm not sure how to decide whether arbitrary sets in R^n are connected. For example, (x,y) in R^2 where at least one of x and y is rational.

Any help would be greatly appreciated
What are you expecting? One nice definition of connectedness is that every continuous $\displaystyle \varphi:X\to\{0,1\}$ where $\displaystyle \{0,1\}$ is the discrete space is constant.

4. Originally Posted by Plato
A set is connected if and only if it cannot be the union of two non-enpty separarted sets.
That is two set neither of which contains a point nor a limit point of the other.
Thanks, but I'm still not sure how to see this for arbitrary sets. For example for the set I posted in the OP, could you tell me whether it is connected or not and why?

Originally Posted by Drexel28
What are you expecting? One nice definition of connectedness is that every continuous $\displaystyle \varphi:X\to\{0,1\}$ where $\displaystyle \{0,1\}$ is the discrete space is constant.
I was just hoping there was a check for connectness that I find easy, like the check for compactness I mentioned.

5. Well I find this is the easiest method:
A set is connected if and only if it cannot be the union of two non-enpty separarted sets.
That is two set neither of which contains a point nor a limit point of the other.
Like for example the set you talked about, lets call it $\displaystyle M$.

If it is not connected then there are two open sets $\displaystyle A,B$
that divide the set.

Now for every vertical line with a rational coordinate, then every point on that line is in $\displaystyle M$
and every point on that line must be in the same set $\displaystyle A$ or $\displaystyle B$. Do you see why?

Now the horizontal lines are the same, for every horizontal line with a rational y-coordinate, the whole line is in $\displaystyle A$ or $\displaystyle B$.

Now pick the vertical line e.g. $\displaystyle (0,y), y\in\mathbb{R}$
and lets assume that it is in $\displaystyle A$, do you see how that leads to every point in the set $\displaystyle M$ is in $\displaystyle A$
so that $\displaystyle M$ is connected?

6. Thanks for the help, that makes sense. As it happens I found a condition for disconnectedness in my notes that makes it a lot easier to me. The statement is: a set T is disconnected if and only if there exists a subset of T (which is not itself or the empty set) that is both open and closed.

For example, for the set I mentioned, this makes it really easy to check. Any non-trivial subset cannot be open (or closed in fact) as for any x in the set, a ball centred at x will contain an infinite amount of points where both coordinates are irrational. Hence the set cannot be disconnected.

7. Originally Posted by Magus01
Thanks for the help, that makes sense. As it happens I found a condition for disconnectedness in my notes that makes it a lot easier to me. The statement is: a set T is disconnected if and only if there exists a subset of T (which is not itself or the empty set) that is both open and closed.

For example, for the set I mentioned, this makes it really easy to check. Any non-trivial subset cannot be open (or closed in fact) as for any x in the set, a ball centred at x will contain an infinite amount of points where both coordinates are irrational. Hence the set cannot be disconnected.
You'll find it's a general theme in topology that proving that a space has a quality is MUCH MUCH more difficult than disproving it. This is of course true in any math subject, though.

8. Originally Posted by Magus01
Any non-trivial subset cannot be open (or closed in fact) as for any x in the set, a ball centred at x will contain an infinite amount of points where both coordinates are irrational. Hence the set cannot be disconnected.
I think you need to consider the subspace topology here. Otherwise you're saying that your original set has no proper open subsets, i.e. that it has the indiscrete topology, which I don't think you wanted.