Showing the sequence of open sets can be written as a union of open sets

Question

Suppose for $\epsilon \gt 0, A \subset \mathbb R $ we have the sequence $(V_{\epsilon}(a))_{a \in A} $

Is it logically correct for me to claim:

$(V_{\epsilon}(a))_{a \in A} $ = $\bigcup_{a \in A} V_{\epsilon}(a)$

Answer 1

On the left hand side you have a sequence of distinct sets, and on the right hand side you have one single set. To claim they are equal is to make a category error.

They are different kinds of things, much like how { {1}, {2}, {3} } is essentially different from {1, 2, 3}.

A related case: The set containing the empty set is not the empty set. This is somewhat different than yours, but similar issues are in play.

Edit: I suppose “category error” is a bit strong, as a sequence can be considered a set. To be as precise as possible, the sequence you give on the left is a set containing sets as elements, while the right is a set containing real numbers as elements.

The right does contain the elements of the sequence as subsets, at least.