Tao Analysis I - 3.5.8

Exercise 3.5.8

Let $X_1, \ldots , X_n$ be sets. Show that the Cartesian product $\prod_{i=1}^{n} X_i$ is empty if and only if at least one of the $X_i$ is empty.

Let's remind ourselves of Definition 3.5.6 for Cartesian products. 

If $(X_i)_{1 \leq i \leq n}$ is an ordered $n$-tuple of sets, we define their Cartesian product as

$$\prod_{1 \leq i \leq n} X_i := \{ (x_i)_{1 \leq i \leq n}: x_i \in X_i \text{ for all } 1 \leq i \leq n \}$$

We need to how both of the following:

• $\prod_{i=1}^{n} X_i = \emptyset \implies \exists X_i = \emptyset$
• $\exists X_i = \emptyset \implies \prod_{i=1}^{n} X_i = \emptyset$

Let's start with the second statement as it is easier to prove. If at least one $X_i$ is empty, let's call it $X_j$ then there is no $x_j= \in X_i$. This means the $n-$tuple $(x_i)_{1 \leq i \leq n}$ can't be formed with $n$ components. Thus the cartesian product, a set, is empty.

Now let's consider the first statement. The cartesian product is empty only if the $n$-tuples can't be formed with $n$ components. This only happens if one of the $X_i$ is empty. 

By proving both statements, we have shown the Cartesian product $\prod_{i=1}^{n} X_i$ is empty if and only if at least one of the $X_i$ is empty. $\square$