Thread: proving a map continuous between topological spaces

I want to show that the function:

f: Z x R -> R define by f(z,x) = x+z

is continuous.

In this case R is endowed with the euclidean topology and Z is the set of integers as a group with the + (normal addition) operator has the discrete topology.

So an open set in U \subset R looks like some union of open balls. The preimage of any such set looks like countably infinitely many copies of these sets shifted by integer amounts.

Now I know that the product topology is the coarsest topology which makes the projection functions continuous.

Can we apply this fact to show the set f^{-1}(U) is in our product topology? How?


Or is it easier than that? Can we show that the preimage of U under f is open because it is the union of open sets in the product topology?

Is it clear that the set {0} x U is open in the product topology? If so then the preimage is the union over z \in Z of {z} x U.

Originally Posted by iknowone

I want to show that the function:

f: Z x R -> R define by f(z,x) = x+z

is continuous.

In this case R is endowed with the euclidean topology and Z is the set of integers as a group with the + (normal addition) operator has the discrete topology.

So an open set in U \subset R looks like some union of open balls. The preimage of any such set looks like countably infinitely many copies of these sets shifted by integer amounts.

Now I know that the product topology is the coarsest topology which makes the projection functions continuous.

Can we apply this fact to show the set f^{-1}(U) is in our product topology? How?


Or is it easier than that? Can we show that the preimage of U under f is open because it is the union of open sets in the product topology?

Is it clear that the set {0} x U is open in the product topology? If so then the preimage is the union over z \in Z of {z} x U.

In your problem an open set of a product topology on has a form , where U is an open set of a discrete topology on and V is an open set of a Euclidean topology on .

Let W be an open set in , which is not equal to itself . We shall show that is open in the product topology on . Since W is an open set in the Euclidean topology on , it is a union of open intervals. Thus, , where . Then, has the form , where , where . Thus, is open in the product topology on .

For the remaining cases,
, which is open in the product topology on .
, which is open in the product topology on .

Thus, f is continuous.