# Thread: Connected spaces

1. ## Connected spaces

I'm having trouble understanding how deleting a point from R2 leaves it a connected space, while deleting a point from R doesn't do so.

Can someone help me with this?

Thanks,
MAX

2. Originally Posted by MAX09
I'm having trouble understanding how deleting a point from R2 leaves it a connected space, while deleting a point from R doesn't do so.
This is one characterization of connect sets:
A set is connected if it is not the union of two non-empty separated sets.
Two sets are separated if neither contains a point or a limit point of the other.

Now $\Re \backslash \{ a\} = \left( { - \infty ,a} \right) \cup \left( {a,\infty } \right)$, clearly the union of two separated sets.

Try that with $\Re \times \Re \backslash \{ (a,b)\}$. What two separated sets would be possible?

3. That clarity about R is quite obvious.. I don't know how I missed it.

Yet, I have one question.
# how do you express RXR/{(a,b)} in terms of set notation?
# Please walk me through the solution...

thanks,
MAX09

4. Originally Posted by MAX09
# how do you express RXR/{(a,b)} in terms of set notation?
There are several different notations.
$\Re ^2 = \Re \times \Re = \left\{ {\left( {x,y} \right):x \in \Re \wedge y \in \Re } \right\}$
So $\Re ^2 \backslash \left\{ {(a,b)} \right\} = \left\{ {\left( {x,y} \right) \in \Re ^2 :x \ne a \wedge y \ne b} \right\}$

5. so, when we delete a point (a,b) from R2 it's the set

R2-(a,b) = {(x,y) such that x !=a & y! =b}

Are we supposed to establish the result using a contradiction?

if a separation existed for the above mentioned set, then...???