Empty Sum, Product, Forall, Exists

Albert Y. C. Lai, trebla [at] vex [dot] net

Notation

For uniformity, I use untraditional notation as exemplified in this table:

notationexpanded
Σi∈{1,2,3}·4×i4×1 + 4×2 + 4×3
Πi∈{1,2,3}·4×i4×1 × 4×2 × 4×3
i∈{1,2,3}·e>ie>1 ∧ e>2 ∧ e>3
i∈{1,2,3}·e>ie>1 ∨ e>2 ∨ e>3

Question

We know and accept that an empty sum like Σi∈{}·4×i is 0. What about empty products like Πi∈{}·4×i?

A similar question concerns ∀ and ∃ statements with empty domains.

Answer

We use this guiding principle: if we split a sum into two sums, or a product into two products, by splitting the index set into two subsets, we should get the same answer. Examples:

Σi∈{1,2,3}·4×i = (Σi∈{1,2}·4×i) + (Σi∈{3}·4×i)
Πi∈{1,2,3}·4×i = (Πi∈{1,2}·4×i) × (Πi∈{3}·4×i)

The splitting may also be so one-sided that one subset receives all the original indexes and the other receives none:

Σi∈{1,2,3}·4×i = (Σi∈{1,2,3}·4×i) + (Σi∈{}·4×i)
Πi∈{1,2,3}·4×i = (Πi∈{1,2,3}·4×i) × (Πi∈{}·4×i)

In order for this to work in general, we need the empty sum and the empty product to satisfy:

y = y + Σi∈{}·4×i
y = y × Πi∈{}·4×i

This forces Σi∈{}·4×i to be 0 and Πi∈{}·4×i to be 1.

Similarly, by considering splitting of domains, ∀ and ∃ statements with empty domains should satisfy:

bb ∧ ∀i∈{}·e>i
bb ∨ ∃i∈{}·e>i

This forces ∀i∈{}·e>i to be true and ∃i∈{}·e>i to be false.


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