How does the product of sets of complex numbers give a character?

Question

I'm working through this "Introduction to Banach Algebras" and just after proposition 8.2 they say:

If $A$ is a commutative Banach algebra, $a\in A$ and $\phi\in M(A)$, then $\phi(a)\in sp(a)$. Hence, $\phi\in\prod_{a\in A}sp(a)$

In this context $\phi$ is a character (which they define as a homomorphism between an algebra and $\mathbb{C}$). $M(A)$ is the set of all C-ideals of A; because every C-ideal maps bijectively to a character with that C-ideal as its kernel, they use $M(A)$ ambiguously to refer to characters or C-ideals.

It makes sense that $\phi(a)\in sp(a)$, but I can't figure out what they mean with the product symbol.

Answer 1

The notation $\prod_{i\in I}S_i$ denotes a set of functions. By definition, $f\in\prod_{i\in I}S_i$ if (i) $f$ is a function with domain $I$ and (ii) $f(i)\in S_i$ for every $i\in I$.

So $\phi\in\prod_{a\in A}sp(a)$. Because $\phi$ is a function with domain $A$ and $\phi(a)\in sp(a)$ for every $a\in A$.

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Come to think of it, that raises an obvious objection that should probably be addressed. Ojection:

"What? $$\prod_{i\in\{0,1\}}A_i=\prod_{i=0}^1A_i=A_0\times A_1.$$That's not a space of functions!"

Well ok, maybe $A_0\times A_1$ is not a space of functions. But it's trivially canonically equivalent to a space of functions. Map the ordered pair $(a_0,a_1)$ to the function $f$ such that $f(0)=a_0$ and $f(1)=a_1$.

Answer 2

The product $\prod_{i\in I}A_i$ of an indexed family of sets is, by definition, the set of all functions $f$ whose domain is the index set $I$ and which satisfy, for each index $i\in I$, the requirement that $f(i)\in A_i$. So the product in your question is the set of functions that assign, to each $a$ in your algebra, an element of its spectrum.

The "product" notation and terminology comes from the fact that, if $I$ and all the sets $A_i$ are finite, then the cardinality of this product is the ordinary, numerical product of the cardinalities of the sets $A_i$.