# Determine all of the open sets in given product topology

1. Feb 18, 2017

### sa1988

1. The problem statement, all variables and given/known data

$X = \{1,2,3\}$ , $\sigma = \big\{\emptyset , \{1,2\}, \{1,2,3\} \big\}$, topology $\{X, \sigma\}$
$Y = \{4,5\}$ , $\tau = \big\{\emptyset , \{4\}, \{4,5\} \big\}$, topology $\{Y, \tau\}$
$Z = \{2,3\} \subset X$

Find all the open sets in the subspace topology on $Z$ and determine all the open sets in the product topology on $Z \times Y$

2. Relevant equations

3. The attempt at a solution

As with previous threads, I'm hoping I'm on the right track. Just looking to check my answers.

Subspace topology on $Z$, $\sigma_z = \big\{\emptyset , \{2\}, \{2, 3\} \big\}$

Product topology is generated by basis $\{ U \times V : U \in \sigma_z , V \in \tau \}$

Thus the product topology on $Z \times Y$ should be created from basis:

$\beta = \big\{\emptyset, \{(2,4)\}, \{(2,4),(2,5)\}, \{(2,4),(2,5),(3,4),(3,5)\}, \{(4,2)\}, \{(4,2),(4,3)\}, \{(4,2),(5,2)\}, \{(4,2),(4,3),(5,2),(5,3)\}\big\}$

Thus all the open sets in the product topology on $Z \times Y$ are found in the union of all sets of basis elements in $\beta$ , plus the empty set:

$\Big\{\big( \cup U_\lambda \big)$ $\forall$ $U_\lambda \subset \beta\Big\} \cup \{\emptyset\}$

Is this all okay? Thanks.

(Many topology threads incoming - this subject is definitely my kryptonite)

2. Feb 18, 2017

### andrewkirk

The subspace topology of $Z$ has a basis with two non-empty sets, and the topology of $Y$ has a basis with two non-empty sets. So the product basis must have $2\times 2=4$ sets. The above list for $\beta$ has seven elements, so cannot be correct. Notice also that the first component of an element of the product set comes from $Z$, which does not contain $4$, yet the above list includes open sets containing elements whose first component is $4$, so they cannot be from $Z\times Y$.

To get the basis of the product topology, just write out the four elements of the Cartesian product of the two bases.

3. Feb 18, 2017

### sa1988

Ah ok, I think I see where I went wrong. For reasons unknown to me, I took the products $Z \times Y$ and $Y \times Z$

Second attempt, I'd go for:

$\beta = \big\{ \{(2,4)\}, \{(2,4),(2,5)\}, \{(2,4), (3,4)\}, \{(2,4),(2,5),(3,4),(3,5)\}\big\}$

And then the open sets in that product topology are all possible unions of all possible subsets of $\beta$, along with the empty set.

4. Feb 18, 2017

### andrewkirk

Yes, that looks right!