Continuity of a function between two topological spaces.

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October 26th 2010, 04:40 PM
okor

Continuity of a function between two topological spaces.

I am have two problems I am having a little trouble with figuring out.

1. f: (X,T)->(Y,D) where D is the discrete topology is always continuous. There are two cases that must be proven: f^-1(empty set) is open on the topology T and f^-1(Y) is open on the topology T. The problem I am having is with f^-1(Y), how can you be certain it is equal to X? I am not sure how to prove this fact.

2. If f: (X,T)->(Y,S) is continuous and A is a subset of X, show that f|A: (A,T)->(Y,S) is continuous. I know since A is a subset of X that for all the open sets in the topological space (A,T) there is some open set V in (Y,S) such that f^-1(V) is equal to that set, but I am not sure how you can be certain that for all open sets V in S that f^-1(V) is still always open.
October 26th 2010, 04:57 PM
HallsofIvy

Quote:

Originally Posted by okor

I am have two problems I am having a little trouble with figuring out.

1. f: (X,T)->(Y,D) where D is the discrete topology is always continuous. There are two cases that must be proven: f^-1(empty set) is open on the topology T and f^-1(Y) is open on the topology T. The problem I am having is with f^-1(Y), how can you be certain it is equal to X? I am not sure how to prove this fact.

By the definition of "f: (X, T)-> (Y, D)" if x is any point in x, then f(x) is in Y. Since "f^-1(A)" is defined as the set of all x such that f(x) is in A, it follows that f^-1(Y)= X. But is "D" the discrete topology or the indiscrete topology? To prove that as function is continuous, you must prove that f^-1(A) is open for every open A. It is the indiscrete topology in which only Y and the empty set are open, NOT the discrete topology.

Quote:

2. If f: (X,T)->(Y,S) is continuous and A is a subset of X, show that f|A: (A,T)->(Y,S) is continuous. I know since A is a subset of X that for all the open sets in the topological space (A,T) there is some open set V in (Y,S) such that f^-1(V) is equal to that set, but I am not sure how you can be certain that for all open sets V in S that f^-1(V) is still always open.

Your statement "for all the open sets in the topological space (A,T) there is some open set V in (Y,S) such that f^-1(V) is equal to that set" is not true. The definition of f:(X,T)->(Y,S) "continuous", for every open set V in (Y,S) f^{-1}(V) is open. It does NOT follow that "given any open set, U, in X, there exist an open set V in (Y, S) so that f^-1(V)= U". For example, let f be the function that maps every point in X into a single point, y, of Y. Then f^-1(V) is either all of X or the empty set, depending upon whethere y is in V or not.

Let f1= f restricted to A. If V is any subset of Y, then [itex]f1^-1(V)= f^-1(V)\cap A[/itex]. If V is open, then f^-1(V) is open so f1^-1(V) is an open set intersect A. What is the definition of "open" in the topology inherited by A?
October 26th 2010, 05:14 PM
okor

Quote:

Originally Posted by HallsofIvy

By the definition of "f: (X, T)-> (Y, D)" if x is any point in x, then f(x) is in Y. Since "f^-1(A)" is defined as the set of all x such that f(x) is in A, it follows that f^-1(Y)= X. But is "D" the discrete topology or the indiscrete topology? To prove that as function is continuous, you must prove that f^-1(A) is open for every open A. It is the indiscrete topology in which only Y and the empty set are open, NOT the discrete topology.

Your statement "for all the open sets in the topological space (A,T) there is some open set V in (Y,S) such that f^-1(V) is equal to that set" is not true. The definition of f:(X,T)->(Y,S) "continuous", for every open set V in (Y,S) f^{-1}(V) is open. It does NOT follow that "given any open set, U, in X, there exist an open set V in (Y, S) so that f^-1(V)= U". For example, let f be the function that maps every point in X into a single point, y, of Y. Then f^-1(V) is either all of X or the empty set, depending upon whethere y is in V or not.

Let f1= f restricted to A. If V is any subset of Y, then [itex]f1^-1(V)= f^-1(V)\cap A[/itex]. If V is open, then f^-1(V) is open so f1^-1(V) is an open set intersect A. What is the definition of "open" in the topology inherited by A?

D'oh! Yeah I did mean the indiscrete topology >.< My apologies.

Thank you for the assistance, the first problem makes a lot more sense now.

For the second problem, an open set in the topology inherited by A is U intersect A where U is open in T. So that means that f1^-1(V) is an open set in T intersected with A so it is open in TA, makes sense! Thank you very much!

Seems so easy now makes me feel kinda dumb :P Thank you for a quick response.