0

Let $(X,\tau)$ be a topological space with a base $\mathfrak{B}$. As $∅ ∈ \tau$, it must be possible: $∃ \beta \subseteq \mathfrak{B} | ∅=\bigcup_{G ∈ \beta}^{}G$. On the other hand, it can be: $\beta = ∅$, because it is true: $∅ \subseteq \mathfrak{B}$, that is: $∅=\bigcup_{G ∈ ∅}^{}G$. Obviously, if $a$ is an element: $a ∈ \bigcup_{G ∈ ∅}^{}G \Leftrightarrow ∃\mathscr{G} ∈ ∅ | a ∈ \mathscr{G}$, that (the right part) is impossible, because not any element can belong to the empty set. So: $a \notin \bigcup_{G ∈ ∅}^{}G$. Now, my question is this: is $\bigcup_{G ∈ ∅}^{}G$ a set? We have seen that $a \notin \bigcup_{G ∈ ∅}^{}G$ not because $\mathscr{G}$ was empty, but because it did not exist, like any $G | G ∈ ∅$. So, can a union of sets that do not exist be possible?

Hanul Jeon
  • 27,376
George
  • 79
  • 4
  • 3
    Typically, for a repeated operation over an empty collection it is taken as definition to be the result such that if you were to extend the operation over a larger set that there be no change. For instance with summation and disjoint sets $A,B$ we would want to have $\sum\limits_{i\in A\sqcup B} i = \sum\limits_{i\in A}i+\sum\limits_{i\in B} i$ and so we have the empty sum $\sum\limits_{i\in \emptyset} i = 0$. Similarly, we would want $\prod\limits_{i\in A\sqcup B} = \prod\limits_{i\in A}i \times \prod\limits_{i\in B} i$ and so we have $\prod\limits_{i\in \emptyset} i = 1$ – JMoravitz Jan 04 '24 at 22:14
  • 1
    Related: https://math.stackexchange.com/questions/384077/does-bigcup-emptyset-equal-bigcup-emptyset/ – Dan Jan 04 '24 at 22:15
  • 3
    An empty union will thusly be the empty set, while the empty intersection will be the universal set (if such a set makes sense in this context). – JMoravitz Jan 04 '24 at 22:15

0 Answers0