#### nate808

this question may sound kind of dumb, so if it is, i apologize, but let say you have a line with a point taken out of it. (<------- -------->) since a point does not have a defined length, how could you prove that the line is now two rays, and not still a line, because it seems a bit contradictory to say that something isn't connected but yet has no distance between it

#### Maxos

It is very simple, the parametrization (Is this word English?) cannot be a continuous (Is this word Maths?) function, then by definition, it is not a curve.

#### Timbuqtu

Topologically a set is called connected if it can not be devided in two disjoint open subsets. Now consider the set A of a line with one point removed from it (for instance $$A = \mathbf{R}- \{ 0 \}$$). We can devide A into two open subsets, namely the two rays (without their starting point). Because they are disjoint, A can not be connected.

#### Maxos

It is incorrect:

$$S \subseteq \mathbb{R}^n$$ is connected $$\leftrightharpoons \exists A, B \subset \mathbb{R}^n : A \cup B = S , A \cap \overline{B} = \varnothing , B \cap \overline{A} = \varnothing$$

#### matt grime

Homework Helper
nate808 said:
this question may sound kind of dumb, so if it is, i apologize, but let say you have a line with a point taken out of it. (<------- -------->) since a point does not have a defined length, how could you prove that the line is now two rays, and not still a line, because it seems a bit contradictory to say that something isn't connected but yet has no distance between it

it is unmathematical and meaningless in the english language to say something "has no distance between it"

look at the definitions anyway (whatever those may be for you).

#### Timbuqtu

Maxos said:
It is incorrect:

$$S \subseteq \mathbb{R}^n$$ is connected $$\leftrightharpoons \exists A, B \subset \mathbb{R}^n : A \cup B = S , A \cap \overline{B} = \varnothing , B \cap \overline{A} = \varnothing$$
Well, yes, that is if you do not require A and B to be open in S. Both statements are equivalent though.
(Recall that A open in S ($$\subset \mathbb{R}^n$$) means that there exist an open $$U \subset \mathbb{R}^n$$ such that $$U \cap S = A$$.)

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