# Thread: proving a map continuous between topological spaces

1. ## 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.

2. 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 $\displaystyle \mathbb{Z} \times \mathbb{Re}$ has a form $\displaystyle U \times V$, where U is an open set of a discrete topology on $\displaystyle \mathbb{Z}$ and V is an open set of a Euclidean topology on $\displaystyle \mathbb{Re}$.

Let W be an open set in $\displaystyle \mathbb{Re}$, which is not equal to $\displaystyle \mathbb{Re}$ itself . We shall show that $\displaystyle f^{-1}(W)$ is open in the product topology on $\displaystyle \mathbb{Z} \times \mathbb{Re}$. Since W is an open set in the Euclidean topology on $\displaystyle \mathbb{Re}$, it is a union of open intervals. Thus, $\displaystyle W=\bigcup_{i} (a_i, b_i)$, where $\displaystyle a_i < b_i$. Then, $\displaystyle f^{-1}(W)$ has the form $\displaystyle \{z\} \times V$, where $\displaystyle V=\bigcup_{i} (a_i-z, b_i-z)$, where $\displaystyle a_i < b_i$. Thus, $\displaystyle f^{-1}(W)$ is open in the product topology on $\displaystyle \mathbb{Z} \times \mathbb{Re}$.

For the remaining cases,
$\displaystyle f^{-1}(\mathbb{Re}) = \mathbb{Z} \times \mathbb{Re}$, which is open in the product topology on $\displaystyle \mathbb{Z} \times \mathbb{Re}$.
$\displaystyle f^{-1}(\emptyset) = \emptyset$, which is open in the product topology on $\displaystyle \mathbb{Z} \times \mathbb{Re}$.

Thus, f is continuous.