Are all functions from a discrete topological space to itself continuous?

[Nusc]

I need to show all fxn f: X -> X are cts in the discrete top. and that the only cts fxns in the concrete top are the csnt fxns.


Let (X,T) be a discrete top with T open sets.

Let f: X->X. WTS that f:X->X is cts if for every open set G in the image of X, f^-1(G) = V is an open in X when V is a subset of X.


Since (X,T) is a discrete top, V in the power set of X must be open. Since T is open and T = PX this implies that f is cts.

Is there anything wrong with that?

As for the xecond half, I'm not sure. When they say cnst fxns do they mean f(c) = c?

 

[HallsofIvy]

Nothing wrong with that. In the discrete topology, every set is open. If f is any function, A any open set (any set) then f^-1(A) is a set and therefore open.

No, f(c)= c is the identity function: f(x)= x. A constant function is of the form f(x)= c for all x in the set.

I'm not sure what you mean by the "concrete" topology. Is that the topology in which the only open sets are X and the empty set? (I would call it the "indiscrete" topology.)

 

[Nusc]

Yes that's the concrete top

 

[HallsofIvy]

Are you sure it's true that the only continuous functions are the constant functions? If f(x)= x, the identity function, then f^-1(empty set)= empty set and f^-1(X)= X. That is, the inverse images of the only open sets are open.