# (path) connected

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• May 11th 2008, 03:05 PM
TXGirl
(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.
• May 11th 2008, 03:26 PM
Plato
There is a standard theorem: Every open connected subset of $R^n$ is path connected.

Now in this case, $R^2$, 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.
• May 11th 2008, 03:35 PM
TheEmptySet
Quote:

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

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

All Euclidean spaces satisfiy the above property.

I hope this helps.
• May 11th 2008, 03:44 PM
TXGirl
Quote:

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

$\mathbb{R}^2$\ $\{ 0,0 \}$
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 $\mathbb{R}^2$ is connected and thus clopen. If so, how is it that there is a subset, namely $\mathbb{R}^2$\ $\{ 0,0 \}$ that is connected, and thus clopen?
• May 11th 2008, 04:28 PM
TheEmptySet
Quote:

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

I think I see my error...

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

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

Since this is the whole space its compliment must be empty

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

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

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

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.
• May 11th 2008, 04:39 PM
TXGirl
Quote:

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

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

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

Since this is the whole space its compliment must be empty

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

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

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

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 $X=\mathbb{R}^2$\ $\{0,0\}$ of $\mathbb{R}^2$

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....
• May 11th 2008, 05:11 PM
Plato
Quote:

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.
• May 11th 2008, 05:38 PM
TXGirl
Quote:

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 $\mathbb{R}^2$\ $\{ 0,0 \}$ 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 $\mathbb{R}^2$ is connected and thus clopen, there is a subset, namely $\mathbb{R}^2$\ $\{ 0,0 \}$, 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 $\mathbb{R}^2$ and $\mathbb{R}^2$\ $\{ 0,0 \}$