1. ## (path) connected

I am having a little trouble with path connectedness.

I am trying to determine whether or not the set R^2\{(0,0)} is path connected.

My first thought was that since R^2\{(0,0)} is a subset of R^2 and R^2, being the cartesian product RxR, is path connected, it follows that R^2\{(0,0)} cannot be both open and closed and thus cannot be path connected.

However, when I imagine it, it seems that given any two non-zero points in R^2, I can connect them by the straight-line path between them (if it does not go through the origin) or I can choose a third point with which to form a broken line connection, perfectly satisfactory as a path.

So that would imply that R^2\{(0,0)} is in fact path connected.

So I think I need clarification of the following iff statement:

A space X is connected iff the only subsets of X that are both open and closed in X are the empty set and X itself.

So, do I understand correctly that this statement says that every path connected set is open and closed? And if so, does this statement not say that no subset of a path connected set can be path connected? It doesn't seem to make sense. What does this statment mean for my space R^2\{(0,0)}?

Any help on the statement as well as the properties (open, closed, connected, path connected) of R^2\{(0,0)} would be greatly appreciated.

2. There is a standard theorem: Every open connected subset of $\displaystyle R^n$ is path connected.

Now in this case, $\displaystyle R^2$, your approach bypasses that theorem.
It is easy to construct a path.
As you note, if the line segment between two points does not contain (0,0) you are done.
It is does we can use a semicircular arc around the origin.

3. Originally Posted by TXGirl
I am having a little trouble with path connectedness.

My first thought was that since R^2\{(0,0)} is a subset of R^2 and R^2, being the cartesian product RxR, is path connected, it follows that R^2\{(0,0)} cannot be both open and closed and thus cannot be path connected.
I think this is what you are thinking of

The only clopen sets (Both open and closed) are the space X and the emptyset.

This is equivelent to X is conncected

$\displaystyle \mathbb{R}^2$\$\displaystyle \{ 0,0 \}$
is a subset of R^2 but is an open set.

All Euclidean spaces satisfiy the above property.

I hope this helps.

4. Originally Posted by TheEmptySet
I think this is what you are thinking of

The only clopen sets (Both open and closed) are the space X and the emptyset.

This is equivelent to X is conncected

$\displaystyle \mathbb{R}^2$\$\displaystyle \{ 0,0 \}$
is a subset of R^2 but is an open set.

All Euclidean spaces satisfiy the above property.

I hope this helps.
So, am I correct that $\displaystyle \mathbb{R}^2$ is connected and thus clopen. If so, how is it that there is a subset, namely $\displaystyle \mathbb{R}^2$\$\displaystyle \{ 0,0 \}$ that is connected, and thus clopen?

5. Originally Posted by TXGirl
So, am I correct that $\displaystyle \mathbb{R}^2$ is connected and thus clopen. If so, how is it that there is a subset, namely $\displaystyle \mathbb{R}^2$\$\displaystyle \{ 0,0 \}$ that is connected, and thus clopen?

I think I see my error...

I was thinking of it as a subspace of $\displaystyle \mathbb{R}^2$

Our Space $\displaystyle X=\mathbb{R}$\$\displaystyle \{0,0\}$

Since this is the whole space its compliment must be empty

$\displaystyle X^c=\emptyset$ also $\displaystyle \emptyset^c=X$

before I claimed that $\displaystyle X^c=\{0,0\}$ I think this is incorrect

X is open because for every point $\displaystyle x \in X$
There exits and $\displaystyle \epsilon > 0$ such that a Ball centered at x of radius epsilon
$\displaystyle B_{\epsilon}(x) \subset X$

The Emptyset is also open

Since they are compliments of each other they are also closed.

I'm sorry if I haven't been very helpful. I guess I don't remember this as well as I thought. Time for more review.

I hope this helps.

Good luck.

6. Originally Posted by TheEmptySet
I think I see my error...

I was thinking of it as a subspace of $\displaystyle \mathbb{R}^2$

Our Space $\displaystyle X=\mathbb{R}$\$\displaystyle \{0,0\}$

Since this is the whole space its compliment must be empty

$\displaystyle X^c=\emptyset$ also $\displaystyle \emptyset^c=X$

before I claimed that $\displaystyle X^c=\{0,0\}$ I think this is incorrect

X is open because for every point $\displaystyle x \in X$
There exits and $\displaystyle \epsilon > 0$ such that a Ball centered at x of radius epsilon
$\displaystyle B_{\epsilon}(x) \subset X$

The Emptyset is also open

Since they are compliments of each other they are also closed.

I'm sorry if I haven't been very helpful. I guess I don't remember this as well as I thought. Time for more review.

I hope this helps.

Good luck.
No, you were correct, we are considering the subspace $\displaystyle X=\mathbb{R}^2$\$\displaystyle \{0,0\}$ of $\displaystyle \mathbb{R}^2$

Any other ideas?? Thanks again for your help. I'm in Germany and we're on holiday until Tuesday, so I cannot go see the prof until then...and for now, I seem to have hit a brick wall in my understanding....

7. Originally Posted by TXGirl
Any other ideas??
Did you even see my reply?
In fact, the other replies have nothing to do with this question.

8. Originally Posted by Plato
Did you even see my reply?
In fact, the other replies have nothing to do with this question.
Yes, I saw it; however, I could see that $\displaystyle \mathbb{R}^2$\$\displaystyle \{ 0,0 \}$ is connected; my problem was, and is, that

there is a theorem saying:

A space X is connected iff the only clopen subsets of X are X and the empty set.

How then is it that if $\displaystyle \mathbb{R}^2$ is connected and thus clopen, there is a subset, namely $\displaystyle \mathbb{R}^2$\$\displaystyle \{ 0,0 \}$, that is connected, and thus clopen?

Somewhere I seem to be misunderstanding the theorem. The way I understand it is seems to say that a subspace of a connected space, cannot be connected; however, this does not seem sensible and, in fact, the problem at hand stands as a contradiction, so I was hoping that someone could explain the theorem, particularly as it relates to the to sets $\displaystyle \mathbb{R}^2$ and $\displaystyle \mathbb{R}^2$\$\displaystyle \{ 0,0 \}$