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If $E$ and $I$ are sets, we use $(E_i)_{i\in I}$ as a notation for a family of parts of $E$ indexed by $I$, i.e. an application $f:I\to \mathcal{P}(E)$, with $E_i:=f(i)$.

In the particular case where $I=\varnothing$, I am wondering if this notation holds (it exists a unique such application $g$, but how would we define $g(i)$ for $i\in\varnothing$...).

I am asking this question to well understand the answer of this post, which deals with a convention about the definition of covering spaces in Hatcher, Algebraic Topology. Thank you for your help !

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Balloon
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    Since there is no $i \in \varnothing$, there is nothing to be defined. $(E_i)_{i \in \varnothing}$ is an empty family of sets. – Daniel Fischer Sep 27 '16 at 21:20
  • Oh right, it is an empty family of sets (i.e. the empty set, and not a family of empty sets or something similar as I was thinking), so indeed all set $U$ in this family is open and mapped homeomorphically to what we want... Thank you for this precision. – Balloon Sep 27 '16 at 21:30
  • @DanielFischer : If I can ask details for one more thing : when I say for all $i\in \varnothing$, let's define $E_i:=f(i),$ it doesn't matter that the RHS doesn't have any signification since I have nothing to define ? – Balloon Sep 27 '16 at 22:11
  • Yes. You do something for every $i\in \varnothing$. Since there is no $i \in \varnothing$, you do … nothing at all. – Daniel Fischer Sep 28 '16 at 09:51
  • @DanielFischer : Okay thank you ! Perhaps a last question if I not annoy you too much : in the answer given in the post, they say that the empty set can be seen as the union of the sets of the family $(E_i){i\in\varnothing},$ that I define as $x\in\cup{i\in\varnothing}E_i\iff\exists i\in\varnothing\text{ s.t. }x\in E_i.$ It seems to well-define the union set, but I can't attach a sense to the assertion $"x\in E_i",$ as $E_i$ is not defined. Should I juste take an other definition in this particular case, or does something is not clear for me (the problem of the assertion not well-defined... – Balloon Sep 28 '16 at 22:13
  • ... appears a second time in the definition of covering space, where Hatcher ask the union $U_\alpha$ to verify that $"\forall \alpha\in A,\exists \cup_{i\in I}E_i \text{ (open disjoint union) s.t. }p^{-1}(U_\alpha)=\cup_{i\in I}E_i \text{ and } \forall i\in\varnothing, E_i\simeq U_\alpha"$). – Balloon Sep 28 '16 at 22:17
  • We don't reach the "$x \in E_i$", since we already stop at "$\exists i \in \varnothing$", as that is false. But, on the other hand, we do have an $E_i$ for every $i \in \varnothing$ (of which there happen to be none). So the statement is well-defined. If you have a little programming experience consider the code int ar[3] = {5,7,16}; if (false) { ar[257] = 3;}. It's perfectly valid. We only do something with the nonexistent object ar[257] if false is true (and if that happens, an out-of-bounds array access is the least of our problems ;-). That's somewhat analogous. – Daniel Fischer Sep 30 '16 at 09:02
  • If the condition is false, what follows needs only be syntactically correct. Any statement of the form $\exists i \in \varnothing \text{ s.t. whatever}$ is vacuously false, like every statement of the form $\forall i \in \varnothing \text{ something}$ is vacuously true. – Daniel Fischer Sep 30 '16 at 09:03
  • @DanielFischer : Really clear, and nice analogy : thank you very much for your explanations ! – Balloon Sep 30 '16 at 11:48

1 Answers1

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The question was answered in the comments above.

Moritz
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