From the first chapter of Sierpinski's "General Topology," discussing Frechet (V)Spaces:
"A set E is connected if and only if it satisfies the following condition W:
If A is a non-empty subset of E with the property that for every element of the set A $\cup$ A' $\cap$ E there exists a neighbourhood U of this element such that U $\cap$ E $\subset$ A, then A = E."
This confuses me because a necessary and sufficient condition for a set E being connected is that it doesn't possess a non-empty proper subset that is both closed and open within E. Condition W says that, in the case E is not connected, it contains a non-empty subset A that is open (because every element of A has a neighbourhood whose elements in E are contained in A), but I don't see how W prevents B = E - A from containing elements of A' $\cap$ E (i.e. A is not closed within E, and therefore E could still be connected).,
It seems like W only says that, if E isn't connected, then A $\cup$ A' $\cap$ E doesn't contain elements of B' $\cap$ E, where B = E - A. But under that condition B could still contain elements of A' $\cap$ E, making W not necessary for showing a set E is connected.
I'm either misreading W, or I have faulty logic.