Tutorial Sheet 6, Topology 2011
... Solution: It is not discrete because {p/q} is not open – if it was {p/q} = U ∩ Q for some open set U ⊂ R. But this isn’t possible - the rational numbers are dense, so any open ball contains infinitely many of them. To see that it is totally disconnected, let C be a component containing two points, x ...
... Solution: It is not discrete because {p/q} is not open – if it was {p/q} = U ∩ Q for some open set U ⊂ R. But this isn’t possible - the rational numbers are dense, so any open ball contains infinitely many of them. To see that it is totally disconnected, let C be a component containing two points, x ...
Finite Topological Spaces - Trace: Tennessee Research and
... Theorem 3.1. Let (X, T ) and (Y, Γ) be topological spaces where X is connected. If f : X → Y is continuous then f (X) is connected. Proof. Suppose to the contrary that {U, V } is a separation of f (X) = Z. Then U and V are each open in the subspace topology of Z. Hence U = Z ∩ Uz and V = Z ∩ Vz wher ...
... Theorem 3.1. Let (X, T ) and (Y, Γ) be topological spaces where X is connected. If f : X → Y is continuous then f (X) is connected. Proof. Suppose to the contrary that {U, V } is a separation of f (X) = Z. Then U and V are each open in the subspace topology of Z. Hence U = Z ∩ Uz and V = Z ∩ Vz wher ...
