Hello,

i have a question about connected and path connected spaces.

in my book i have read, that every path-connected space is also connected.

It is proved there as follows:

Suppose that X isn't connected then there exist nonempty-open sets A,B s.t

X=A $\displaystyle \cup $ B.

We have a continuous path g:I->X with I=$\displaystyle g^-1(A) \cup g^-1(B)$

But I is connected, so here is a contradiction.

I want to know, why this proof is correct? I mean I is sometimes open sometimes not determined by the topology given on I. So how can we say I is open.

I think the author means the intervall I with the standard-topology given by |.|.

But why does this proof works? I'm a little bit confused.

I hope someone can explain this problem for me.